Correct Answer: malloc()
Explanation: The `malloc()` function is used to allocate a specified number of bytes of memory during the execution of a program.
Correct Answer: calloc()
Explanation: The `calloc()` function allocates memory for an array of elements, initializes them to zero, and then returns a pointer to the memory.
Correct Answer: Reallocate memory
Explanation: The `realloc()` function changes the size of previously allocated memory block to a new size.
Correct Answer: free()
Explanation: The `free()` function is used to release memory that was previously allocated using `malloc()`, `calloc()`, or `realloc()`.
Correct Answer: It may cause undefined behavior
Explanation: Freeing already freed memory can lead to undefined behavior, including crashes or data corruption.
Correct Answer: <stdlib.h>
Explanation: The <stdlib.h> header file contains the declarations for memory allocation functions such as `malloc()`, `calloc()`, `realloc()`, and `free()`.
Correct Answer: void*
Explanation: The `malloc()` function returns a pointer of type `void*` which can be cast to any other pointer type.
Correct Answer: NULL
Explanation: When `malloc()` fails to allocate the requested memory, it returns NULL to indicate the failure.
Correct Answer: Using `realloc()`
Explanation: The `realloc()` function is used to resize a previously allocated memory block to a new size.
Correct Answer: calloc()
Explanation: The `calloc()` function allocates memory for an array and initializes all its bits to zero.
Correct Answer: Allocates memory for 10 integers
Explanation: The `malloc` function allocates a block of memory. `sizeof(int) * 10` calculates the amount of memory needed for 10 integers. The `(int*)` cast converts the void pointer returned by `malloc` to an integer pointer. This way, `ptr` can be used as an array of 10 integers.
Correct Answer: `malloc` will return NULL
Explanation: If `malloc` is unable to allocate the requested memory, it returns a NULL pointer. This indicates that the memory allocation failed. It is good practice to always check if the returned pointer is NULL before using it to avoid dereferencing a NULL pointer, which can lead to program crashes.
Correct Answer: Deallocates memory pointed to by `ptr`
Explanation: The `free` function deallocates the memory that was previously allocated by `malloc`, `calloc`, or `realloc`. After calling `free`, the pointer `ptr` becomes a dangling pointer, meaning it points to a memory location that is no longer valid. It is a good practice to set the pointer to NULL after freeing it to avoid accidental usage.
Correct Answer: To prevent memory leaks
Explanation: Dynamically allocated memory that is not freed results in a memory leak. A memory leak occurs when a program loses the ability to free the memory that is no longer needed, which can lead to increased memory usage and eventually cause the program or system to run out of memory. This is especially critical in long-running applications or systems with limited memory resources.
Correct Answer: `malloc` allocates memory without initializing, while `calloc` allocates and initializes memory to zero
Explanation: `malloc` allocates a block of memory but does not initialize it, meaning it contains garbage values. In contrast, `calloc` allocates memory and initializes all bits to zero. `calloc` is often used when initializing arrays or buffers where a zero-initialized block of memory is needed.
Correct Answer: When allocated memory is not accessible
Explanation: A memory leak occurs when a program allocates memory dynamically but fails to deallocate it before losing the reference to that memory. This results in the allocated memory being inaccessible and unusable for the program, effectively reducing the amount of available memory. Over time, memory leaks can exhaust the available memory, leading to performance degradation or program crashes.
Correct Answer: To determine the amount of memory to allocate
Explanation: The `sizeof` operator is used to determine the size in bytes of a data type or object. When using `malloc` or `calloc` to allocate memory, `sizeof` ensures that the correct amount of memory is allocated for the specified data type. For example, `sizeof(int)` returns the size of an integer in bytes, ensuring that the allocated memory block is large enough to store the intended data.
Correct Answer: `realloc` can both increase and decrease the size of the allocated memory
Explanation: The `realloc` function can be used to resize a previously allocated memory block. It can increase the size of the memory block if more space is needed or decrease the size if less space is required. If the new size is larger, `realloc` may move the memory block to a new location if there is not enough contiguous space at the current location, copying the existing data to the new block.
Correct Answer: It is automatically freed
Explanation: When `realloc` moves the memory block to a new location, the original memory block is automatically freed. This ensures that there are no memory leaks when resizing memory. However, if `realloc` fails to allocate the new block, the original block remains unchanged, and `realloc` returns NULL, allowing the program to handle the allocation failure without losing the original data.
Correct Answer: Set the pointer to NULL
Explanation: After calling `free` on a dynamically allocated memory block, it is a good practice to set the pointer to NULL. This prevents the pointer from becoming a dangling pointer, which points to a memory location that has been deallocated. Accessing a dangling pointer can lead to undefined behavior, including program crashes and data corruption. Setting the pointer to NULL ensures that any accidental dereference can be detected easily.
Correct Answer: calloc()
Explanation: The `calloc` function is often used for allocating memory for arrays, including 2D arrays, because it initializes the allocated memory to zero. This is particularly useful when you need to ensure that all elements in the array are initialized to zero before using them. `calloc` takes two arguments: the number of elements and the size of each element.
Correct Answer: char *str = (char*)malloc(50 * sizeof(char));
Explanation: To allocate memory for a string of 50 characters, you need to allocate 50 bytes (assuming 1 byte per character). The expression `50 * sizeof(char)` ensures that the memory allocation is correctly sized for 50 characters. The `malloc` function allocates the required memory, and the `(char*)` cast converts the void pointer returned by `malloc` to a char pointer.
Correct Answer: The program will run normally
Explanation: Calling `free` on a NULL pointer is safe and has no effect. The C standard specifies that if the pointer passed to `free` is NULL, no action occurs. This makes it safe to `free` pointers that might be NULL without needing to check if they are NULL first.
Correct Answer: realloc()
Explanation: The `realloc` function is used to resize a previously allocated memory block. It can increase or decrease the size of the allocated memory. This is useful when you initially allocate a memory block with `malloc` or `calloc` and later determine that you need more or less memory.
Correct Answer: Allocates memory for 10 integers
Explanation: The `realloc` function can also be used to allocate memory when the first argument is NULL. This behavior is equivalent to calling `malloc`. In this case, `realloc(NULL, 10 * sizeof(int))` allocates memory for 10 integers, similar to `malloc(10 * sizeof(int))`.
Correct Answer: If the returned pointer is NULL
Explanation: After calling `malloc` or `calloc`, it is crucial to check if the returned pointer is NULL. A NULL pointer indicates that the memory allocation failed, which can happen if there is not enough memory available. Failing to check for a NULL pointer and proceeding to use it can lead to dereferencing a NULL pointer, causing the program to crash.
Correct Answer: struct MyStruct *ptr = (struct MyStruct*)malloc(sizeof(struct MyStruct));
Explanation: When allocating memory for a structure, you use the `sizeof` operator to determine the size of the structure. The `malloc` function allocates the required memory, and the `(struct MyStruct*)` cast converts the void pointer returned by `malloc` to a pointer of the correct type. This ensures that the pointer can be used to access the structure’s members correctly.
Correct Answer: float *arr = (float*)calloc(10, sizeof(float));
Explanation: The `calloc` function is used to allocate memory for an array and initialize all its elements to zero. The first argument specifies the number of elements, and the second argument specifies the size of each element. In this case, `calloc(10, sizeof(float))` allocates memory for 10 floats and initializes all elements to zero.
Correct Answer: Undefined behavior occurs
Explanation: Using `free` on a memory block that was not dynamically allocated (e.g., a stack-allocated variable) results in undefined behavior. This can cause the program to crash, corrupt data, or exhibit other unexpected behavior. It is essential to only `free` memory that was allocated dynamically using `malloc`, `calloc`, or `realloc`.
Correct Answer: Data beyond the new size is lost
Explanation: When using `realloc` to shrink the size of a memory block, any data stored beyond the new size is lost. `realloc` may move the memory block to a new location and copy only the data that fits within the new size. It is essential to ensure that no important data exists beyond the new size before shrinking the memory block to avoid data loss.
Correct Answer: Allocating memory and not freeing it
Explanation: A memory leak occurs when a program allocates memory dynamically using functions like `malloc`, `calloc`, or `realloc` and fails to free that memory using `free` when it is no longer needed. Over time, this can lead to a depletion of available memory, which can cause the program or system to run out of memory and potentially crash.
Correct Answer: Freeing dynamically allocated memory when it is no longer needed
Explanation: To prevent memory leaks, it is crucial to free dynamically allocated memory once it is no longer required. This can be done using the `free` function. Failing to do so keeps the memory allocated and inaccessible for future use, leading to memory leaks.
Correct Answer: They detect memory leaks and memory usage errors
Explanation: Tools like Valgrind are used to detect memory leaks, memory usage errors, and other memory-related issues in C programs. Valgrind can help identify places in the code where memory is allocated but not freed, where memory is accessed after being freed, and other common mistakes that lead to memory leaks and bugs.
Correct Answer: Gradually increasing memory usage over time
Explanation: A common sign of a memory leak is that the memory usage of the program gradually increases over time, even if the workload remains constant. This happens because the program is continuously allocating memory without freeing it, leading to a steady increase in memory consumption.
Correct Answer: By automatically freeing memory when it is no longer needed
Explanation: Smart pointers in C++ (such as `std::unique_ptr` and `std::shared_ptr`) automatically manage the lifetime of dynamically allocated memory. When a smart pointer goes out of scope, it automatically frees the memory it points to, preventing memory leaks. This automatic management reduces the risk of forgetting to free memory manually.
Correct Answer: To execute cleanup functions at program termination
Explanation: The `atexit` function in C allows you to register one or more functions to be called when the program terminates. These functions can perform necessary cleanup tasks, such as freeing dynamically allocated memory, closing files, and other resource management activities. This helps ensure that resources are properly released even if the program exits unexpectedly.
Correct Answer: To avoid dangling pointer dereferencing
Explanation: After freeing the memory pointed to by a pointer, setting the pointer to NULL helps prevent accidental dereferencing of the now-invalid pointer. Dereferencing a dangling pointer (a pointer that points to freed memory) can lead to undefined behavior, such as program crashes and data corruption. Setting the pointer to NULL ensures that any attempts to use it will result in a null pointer dereference, which is easier to detect and debug.
Correct Answer: By making memory inaccessible for deallocation
Explanation: Circular references occur when two or more objects reference each other, forming a cycle. In garbage-collected languages, this can prevent the memory from being freed, as the reference count of the objects never reaches zero. In C, circular references can occur with custom memory management schemes, making it difficult to determine when it is safe to free the memory, potentially leading to memory leaks.
Correct Answer: Failing to free allocated memory
Explanation: The primary cause of memory leaks in programs that use dynamic memory allocation is failing to free allocated memory when it is no longer needed. This oversight leads to memory that is no longer accessible but still allocated, reducing the amount of available memory and potentially causing the program or system to run out of memory.
Correct Answer: By using error handling techniques like `goto` for cleanup
Explanation: One way to ensure that dynamically allocated memory is always freed, even if an error occurs, is to use error handling techniques like `goto` to jump to a cleanup section of the code. This section can contain code to free all allocated memory and perform other cleanup tasks. Using `goto` for cleanup helps centralize the cleanup code and ensures that resources are properly released in case of errors.
Correct Answer: malloc()
Explanation: The `malloc` function allocates memory but does not initialize it. The allocated memory contains garbage values. If you need the memory to be initialized to zero, you should use `calloc`, which allocates and initializes the memory to zero.
Correct Answer: Uninitialized (garbage values)
Explanation: Memory allocated by `malloc` is not initialized, which means it contains garbage values. The content of the memory will be whatever was present at that location before the allocation. If initialization to zero is needed, `calloc` should be used instead.
Correct Answer: By freeing memory in the reverse order of allocation
Explanation: Memory fragmentation can be minimized by freeing memory in the reverse order of allocation, known as LIFO (Last In, First Out). This helps keep large contiguous blocks of free memory available, reducing fragmentation. Fragmentation occurs when there are many small, non-contiguous free memory blocks, making it difficult to allocate large blocks of memory.
Correct Answer: memcpy()
Explanation: The `memcpy` function is used to copy a specified number of bytes from one memory location to another. It takes three arguments: the destination pointer, the source pointer, and the number of bytes to copy. It is useful for copying data between dynamically allocated memory blocks.
Correct Answer: NULL
Explanation: If the new size specified in `realloc` is zero, the function frees the memory block and returns NULL. This effectively acts as a call to `free` for the given pointer. It is important to handle this case to avoid dereferencing a NULL pointer.
Correct Answer: It behaves like `malloc`
Explanation: When `realloc` is called with a NULL pointer, it behaves like `malloc`. It allocates a new block of memory of the specified size and returns a pointer to it. This is useful for initially allocating memory and resizing it later if needed.
Correct Answer: To deallocate previously allocated memory
Explanation: The `free` function is used to deallocate memory that was previously allocated by `malloc`, `calloc`, or `realloc`. It releases the memory back to the system so that it can be reused by other parts of the program or other programs. Failing to use `free` can result in memory leaks.
Correct Answer: By setting the pointer to NULL after `free`
Explanation: Double freeing occurs when `free` is called more than once on the same pointer, which can lead to undefined behavior and program crashes. To avoid this, it is good practice to set the pointer to NULL immediately after freeing it. This ensures that any subsequent calls to `free` on that pointer will have no effect.
Correct Answer: It determines the size of memory to allocate
Explanation: The `sizeof` operator is used to determine the size in bytes of a data type or object. This is essential in dynamic memory allocation to ensure that the correct amount of memory is allocated. For example, `sizeof(int)` returns the size of an integer in bytes, which can be used to allocate memory for an array of integers.
Correct Answer: calloc()
Explanation: The `calloc` function allocates memory for an array of a specified number of elements and initializes all elements to zero. For example, `calloc(100, sizeof(int))` allocates memory for 100 integers and initializes each integer to zero. This is useful for ensuring that all elements in the array are initially set to a known value.
int *arr;
arr = (int*)malloc(10 * sizeof(int));
if (arr != NULL) {
for (int i = 0; i < 10; i++) {
arr[i] = i;
}
}
Correct Answer: Allocates memory for an array of 10 integers and initializes them to their index values
Explanation: The code allocates memory for an array of 10 integers using `malloc`. It then checks if the allocation was successful and initializes each element of the array to its index value using a for loop.
char *str;
str = (char*)malloc(20 * sizeof(char));
if (str == NULL) {
printf("Memory allocation failed\n");
} else {
strcpy(str, "Hello, World!");
printf("%s\n", str);
}
free(str);
Correct Answer: Memory allocation failed
Explanation: If `malloc` fails to allocate memory, it returns NULL. The code checks if `str` is NULL and prints "Memory allocation failed" if it is. Otherwise, it copies "Hello, World!" into `str` and prints it. Since the question specifies the scenario where `malloc` fails, the output will be "Memory allocation failed".
Correct Answer: `int *arr = (int*)realloc(arr, 10 * sizeof(int));`
Explanation: The `realloc` function is used to resize an allocated memory block. The correct way to reallocate memory for an array of 5 integers to hold 10 integers is `realloc(arr, 10 * sizeof(int))`, which adjusts the size of the memory block to accommodate 10 integers.
int *arr;
arr = (int*)malloc(5 * sizeof(int));
arr[5] = 10;
Correct Answer: Undefined behavior
Explanation: Accessing an element beyond the allocated memory (in this case, `arr[5]`) leads to undefined behavior. The memory allocated by `malloc` is only guaranteed to be accessible within the specified size. Accessing outside of this range can result in program crashes, data corruption, or other unpredictable behavior.
float *ptr;
ptr = (float*)malloc(5 * sizeof(float));
/* Some code that uses ptr */
free(ptr);
Correct Answer: `free(ptr);`
Explanation: The `free` function is used to deallocate memory that was previously allocated with `malloc`, `calloc`, or `realloc`. The correct way to free the memory pointed to by `ptr` is `free(ptr);`. This releases the memory back to the system.
int *p = (int*)malloc(sizeof(int));
if (p == NULL) {
printf("Memory allocation failed\n");
} else {
*p = 42;
printf("%d\n", *p);
}
free(p);
Correct Answer: 42
Explanation: The code allocates memory for a single integer and assigns it the value 42. It then prints the value pointed to by `p`, which is 42. If the memory allocation fails, it prints "Memory allocation failed". Since the allocation is assumed to be successful in this scenario, the output will be 42.
int *arr;
arr = (int*)malloc(5 * sizeof(int));
free(arr);
free(arr);
Correct Answer: The program will cause undefined behavior
Explanation: Calling `free` on a pointer more than once (double free) results in undefined behavior. This can lead to program crashes, memory corruption, and other unpredictable issues. After freeing a pointer, it should not be freed again.
int *ptr;
ptr = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
printf("%d ", ptr[i]);
}
free(ptr);
Correct Answer: Prints five zeros
Explanation: The `calloc` function allocates memory for an array of 5 integers and initializes all elements to zero. The for loop prints each element of the array, resulting in five zeros being printed. The memory is then freed using `free`.
int *arr1 = (int*)malloc(5 * sizeof(int));
int *arr2 = (int*)realloc(arr1, 10 * sizeof(int));
Correct Answer: Resizing the memory block for `arr1` to hold 10 integers
Explanation: The code snippet shows the use of `realloc` to resize the memory block pointed to by `arr1` from 5 integers to 10 integers. The `realloc` function adjusts the size of the previously allocated memory block and returns a pointer to the new memory block.
Correct Answer: Memory allocated with `malloc` must be initialized manually
Explanation: Memory allocated with `malloc` is not initialized and contains garbage values. It must be initialized manually if initialization is needed. In contrast, `calloc` initializes the allocated memory to zero. Memory allocated with `realloc` can be resized, and dynamically allocated memory must be explicitly freed using `free`; it is not automatically freed at program termination.
int *arr;
arr = (int*)malloc(3 * sizeof(int));
arr[0] = 1;
arr[1] = 2;
arr[2] = 3;
free(arr);
printf("%d\n", arr[1]);
Correct Answer: Undefined behavior
Explanation: After calling `free`, the memory pointed to by `arr` is deallocated, and accessing it results in undefined behavior. Therefore, the code may print any value or crash.
int *p;
p = (int*)malloc(sizeof(int));
if (p == NULL) {
printf("Memory allocation failed\n");
return 1;
}
*p = 100;
printf("%d\n", *p);
free(p);
Correct Answer: 100
Explanation: The code allocates memory for an integer, assigns the value 100 to it, and prints this value. Since memory allocation is assumed to be successful, the output will be 100.
int *ptr;
ptr = (int*)calloc(4, sizeof(int));
if (ptr != NULL) {
for (int i = 0; i < 4; i++) {
printf("%d ", ptr[i]);
}
free(ptr);
}
Correct Answer: Prints four zeros
Explanation: The `calloc` function allocates memory for an array of 4 integers and initializes all elements to zero. The for loop prints each element of the array, resulting in four zeros being printed.
int *a, *b;
a = (int*)malloc(5 * sizeof(int));
b = a;
free(a);
free(b);
Correct Answer: Double free error
Explanation: Both `a` and `b` point to the same allocated memory. Calling `free` on both `a` and `b` results in a double free error, which leads to undefined behavior, potentially causing the program to crash.
char *str;
str = (char*)malloc(6 * sizeof(char));
strcpy(str, "Hello");
free(str);
printf("%s\n", str);
Correct Answer: Undefined behavior
Explanation: After freeing `str`, the memory is deallocated, and accessing it (as in `printf("%s\n", str)`) results in undefined behavior. The program may print any value or crash.
int *p = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
p[i] = i * i;
}
p = (int*)realloc(p, 10 * sizeof(int));
for (int i = 5; i < 10; i++) {
p[i] = i * i;
}
for (int i = 0; i < 10; i++) {
printf("%d ", p[i]);
}
free(p);
Correct Answer: Allocates memory for 5 integers, then reallocates it to 10 integers and initializes with squares of indices
Explanation: The code first allocates memory for 5 integers and initializes them to the squares of their indices. It then reallocates memory to hold 10 integers, initializes the new elements to the squares of their indices, and prints the entire array.
int *arr;
arr = (int*)malloc(3 * sizeof(int));
if (arr == NULL) {
printf("Memory allocation failed\n");
return 1;
}
arr[0] = 10;
arr[1] = 20;
arr[2] = 30;
free(arr);
arr = (int*)malloc(3 * sizeof(int));
printf("%d\n", arr[1]);
free(arr);
Correct Answer: Undefined behavior
Explanation: After freeing `arr`, allocating memory again for `arr` does not guarantee that it will point to the same memory block. Accessing `arr[1]` after reallocation results in undefined behavior because the new memory block is uninitialized.
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i;
}
int *new_arr = (int*)realloc(arr, 10 * sizeof(int));
if (new_arr == NULL) {
printf("Reallocation failed\n");
free(arr);
return 1;
}
for (int i = 5; i < 10; i++) {
new_arr[i] = i;
}
for (int i = 0; i < 10; i++) {
printf("%d ", new_arr[i]);
}
free(new_arr);
Correct Answer: Reallocation failed
Explanation: If `realloc` fails, it returns NULL and does not free the original memory block. The code checks if `new_arr` is NULL and prints "Reallocation failed" if it is. Since the question specifies the scenario where `realloc` fails, the output will be "Reallocation failed".
int *arr;
arr = (int*)calloc(5, sizeof(int));
if (arr != NULL) {
for (int i = 0; i < 5; i++) {
printf("%d ", arr[i]);
}
free(arr);
}
Correct Answer: Prints five zeros
Explanation: The `calloc` function allocates memory for an array of 5 integers and initializes all elements to zero. The for loop prints each element of the array, resulting in five zeros being printed. The memory is then freed using `free`.
int *arr = (int*)malloc(3 * sizeof(int));
arr[0] = 1;
arr[1] = 2;
arr[2] = 3;
free(arr);
arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = arr[i] + 1;
}
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: Undefined behavior
Explanation: After calling `free`, the memory pointed to by `arr` is deallocated. The subsequent `malloc` call allocates new memory, which is uninitialized. The code attempts to access and modify the uninitialized values, leading to undefined behavior.
int *ptr = NULL;
ptr = (int*)malloc(4 * sizeof(int));
if (ptr != NULL) {
for (int i = 0; i < 4; i++) {
ptr[i] = i * 10;
}
free(ptr);
ptr = NULL;
}
Correct Answer: Allocates memory for 4 integers, initializes them, frees the memory, and sets pointer to NULL
Explanation: The code allocates memory for 4 integers and initializes each element to `i * 10`. After that, it frees the allocated memory and sets the pointer to NULL, which is a good practice to avoid dangling pointers.
int *arr = (int*)malloc(2 * sizeof(int));
arr[0] = 5;
arr[1] = 10;
arr = (int*)realloc(arr, 3 * sizeof(int));
arr[2] = 15;
printf("%d %d %d\n", arr[0], arr[1], arr[2]);
free(arr);
Correct Answer: 5 10 15
Explanation: The `realloc` function resizes the allocated memory to hold 3 integers. The existing values are preserved, and the new element is initialized to 15. The code prints 5, 10, and 15 as expected.
char *str = (char*)malloc(6 * sizeof(char));
strcpy(str, "Hello");
str = (char*)realloc(str, 12 * sizeof(char));
strcat(str, " World");
printf("%s\n", str);
free(str);
Correct Answer: Hello World
Explanation: The `realloc` function resizes the allocated memory to hold 12 characters. The existing string "Hello" is preserved, and " World" is concatenated to it. The code prints "Hello World".
int *arr = (int*)malloc(3 * sizeof(int));
arr[0] = 1;
arr[1] = 2;
arr[2] = 3;
free(arr);
arr = NULL;
printf("%d\n", arr == NULL);
Correct Answer: 1
Explanation: After freeing the allocated memory, the pointer `arr` is set to NULL. The `printf` statement checks if `arr` is NULL, which is true, so it prints 1.
int *a = (int*)malloc(5 * sizeof(int));
if (a == NULL) {
printf("Memory allocation failed\n");
return 1;
}
int *b = (int*)malloc(5 * sizeof(int));
if (b == NULL) {
printf("Memory allocation failed\n");
free(a);
return 1;
}
memcpy(b, a, 5 * sizeof(int));
free(a);
free(b);
Correct Answer: Allocates memory for two arrays and copies contents from `a` to `b`
Explanation: The code allocates memory for two arrays of 5 integers each. It uses `memcpy` to copy the contents of `a` to `b`, then frees both memory blocks. If any allocation fails, it handles the failure gracefully by freeing the previously allocated memory.
int *p;
p = (int*)calloc(5, sizeof(int));
if (p != NULL) {
for (int i = 0; i < 5; i++) {
printf("%d ", p[i]);
}
free(p);
}
Correct Answer: Allocates memory for 5 integers, initializes them to zero, and prints them
Explanation: The `calloc` function allocates memory for 5 integers and initializes all elements to zero. The for loop prints each element of the array, which results in five zeros being printed. The memory is then freed using `free`.
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i + 1;
}
free(arr);
arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = (i + 1) * 2;
}
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: 2 4 6
Explanation: The code first allocates memory for 5 integers and initializes them. After freeing this memory, it allocates memory for 3 integers and initializes these new elements with values 2, 4, and 6. The code prints 2, 4, 6 and then frees the allocated memory.
int *ptr = (int*)malloc(5 * sizeof(int));
if (ptr != NULL) {
for (int i = 0; i < 5; i++) {
ptr[i] = i;
}
free(ptr);
}
ptr = (int*)calloc(5, sizeof(int));
if (ptr != NULL) {
for (int i = 0; i < 5; i++) {
printf("%d ", ptr[i]);
}
free(ptr);
}
Correct Answer: Allocates memory for 5 integers, initializes with index values, frees, allocates again, and prints zeros
Explanation: The code allocates memory for 5 integers, initializes them with their index values, and frees the memory. It then allocates memory again using `calloc`, which initializes all elements to zero, and prints these values.
int *a = (int*)malloc(4 * sizeof(int));
for (int i = 0; i < 4; i++) {
a[i] = i + 1;
}
int *b = (int*)malloc(4 * sizeof(int));
memcpy(b, a, 4 * sizeof(int));
for (int i = 0; i < 4; i++) {
printf("%d ", b[i]);
}
free(a);
free(b);
Correct Answer: 1 2 3 4
Explanation: The code allocates memory for two arrays of 4 integers each. It initializes the first array and copies its contents to the second array using `memcpy`. The code prints the elements of the second array, which are 1, 2, 3, and 4, and then frees both memory blocks.
char *str = (char*)malloc(5 * sizeof(char));
strcpy(str, "Test");
str = (char*)realloc(str, 10 * sizeof(char));
strcat(str, "ing");
printf("%s\n", str);
free(str);
Correct Answer: Testing
Explanation: The code allocates memory for a string, copies "Test" into it, reallocates memory to expand the string, concatenates "ing" to the existing string, and prints "Testing". The memory is then freed.
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i * 2;
}
free(arr);
arr = (int*)calloc(3, sizeof(int));
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: 0 0 0
Explanation: The code first allocates memory for 3 integers and initializes them to 0, 2, and 4. After freeing this memory, it allocates memory again using `calloc`, which initializes all elements to zero. The code prints three zeros.
int *ptr = (int*)malloc(2 * sizeof(int));
ptr[0] = 5;
ptr[1] = 10;
ptr = (int*)realloc(ptr, 3 * sizeof(int));
ptr[2] = 15;
for (int i = 0; i < 3; i++) {
printf("%d ", ptr[i]);
}
free(ptr);
Correct Answer: 5 10 15
Explanation: The `realloc` function resizes the allocated memory to hold 3 integers. The existing values are preserved, and the new element is initialized to 15. The code prints 5, 10, and 15 as expected.
int *arr = (int*)malloc(4 * sizeof(int));
for (int i = 0; i < 4; i++) {
arr[i] = i;
}
free(arr);
arr = (int*)malloc(4 * sizeof(int));
for (int i = 0; i < 4; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: Prints four uninitialized values
Explanation: After the first `malloc` and initialization, the memory is freed. The second `malloc` allocates new memory, which is uninitialized. The code prints the uninitialized values of the new memory block.
int *ptr = (int*)malloc(2 * sizeof(int));
ptr[0] = 1;
ptr[1] = 2;
free(ptr);
ptr = (int*)malloc(2 * sizeof(int));
ptr[0] = 3;
ptr[1] = 4;
printf("%d %d\n", ptr[0], ptr[1]);
free(ptr);
Correct Answer: 3 4
Explanation: The code allocates memory for 2 integers, initializes them, and then frees this memory. It then allocates new memory for 2 integers again and initializes these new elements to 3 and 4. The code prints 3 and 4.
int *arr = (int*)calloc(3, sizeof(int));
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: Allocates memory for 3 integers, initializes them to zero, and prints them
Explanation: The `calloc` function allocates memory for 3 integers and initializes all elements to zero. The for loop prints each element of the array, which results in three zeros being printed. The memory is then freed using `free`.
int *a = (int*)malloc(3 * sizeof(int));
a[0] = 10;
a[1] = 20;
a[2] = 30;
int *b = (int*)malloc(3 * sizeof(int));
memcpy(b, a, 3 * sizeof(int));
for (int i = 0; i < 3; i++) {
printf("%d ", b[i]);
}
free(a);
free(b);
Correct Answer: 10 20 30
Explanation: The code allocates memory for two arrays of 3 integers each. It initializes the first array and copies its contents to the second array using `memcpy`. The code prints the elements of the second array, which are 10, 20, and 30, and then frees both memory blocks.
int *arr = (int*)malloc(3 * sizeof(int));
arr[0] = 1;
arr[1] = 2;
arr[2] = 3;
free(arr);
arr = (int*)realloc(arr, 6 * sizeof(int));
for (int i = 0; i < 6; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: Undefined behavior
Explanation: After `free(arr)`, the memory is deallocated, and accessing it (even with `realloc`) results in undefined behavior. The new allocation may or may not preserve the original values and will contain uninitialized values for the new elements.
char *str;
str = (char*)malloc(6 * sizeof(char));
strcpy(str, "Hello");
free(str);
str = (char*)calloc(6, sizeof(char));
printf("%s\n", str);
free(str);
Correct Answer: (empty string)
Explanation: The code allocates memory for a string and copies "Hello" into it, then frees this memory. It then allocates new memory using `calloc`, which initializes all elements to zero. The `printf` statement prints an empty string as the new memory block is initialized to zero.
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i * 5;
}
free(arr);
arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
free(arr);
Correct Answer: Prints three uninitialized values
Explanation: After the first `malloc` and initialization, the memory is freed. The second `malloc` allocates new memory, which is uninitialized. The code prints the uninitialized values of the new memory block.
int *a = (int*)malloc(3 * sizeof(int));
a[0] = 1;
a[1] = 2;
a[2] = 3;
int *b = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
b[i] = a[i] * 2;
}
for (int i = 0; i < 3; i++) {
printf("%d ", b[i]);
}
free(a);
free(b);
Correct Answer: 2 4 6
Explanation: The code allocates memory for two arrays of 3 integers each. It initializes the first array with values 1, 2, and 3. It then initializes the second array with values that are double the corresponding values of the first array. The code prints 2, 4, and 6 and then frees both memory blocks.
int *arr = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = 0;
}
int *arr = (int*)calloc(5, sizeof(int));
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = 0;
}
int *arr = (int*)malloc(5 * sizeof(int));
Correct Answer: .
Explanation: The `calloc` function automatically initializes all allocated memory to zero. Option B correctly uses `calloc` to allocate memory for 5 integers and initializes them to zero.
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
arr = (int*)realloc(arr, 5 * sizeof(int));
for (int i = 3; i < 5; i++) {
arr[i] = 0;
}
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
arr = (int*)realloc(arr, 5 * sizeof(int));
int *arr = (int*)realloc(NULL, 5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = 0;
}
int *arr = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
Correct Answer: .
Explanation: Option A correctly resizes the array using `realloc` and then initializes the new elements to zero. `realloc` preserves the existing values and does not initialize the new elements, so they must be explicitly set to zero.
int *arr = (int*)malloc(5 * sizeof(int));
free(arr);
int *arr = (int*)malloc(5 * sizeof(int));
free(arr);
arr = NULL;
int *arr = (int*)calloc(5, sizeof(int));
free(arr);
arr = NULL;
int *arr = (int*)calloc(5, sizeof(int));
free(arr);
Correct Answer: .
Explanation: Option B correctly frees the allocated memory and then sets the pointer to NULL. This practice helps avoid dangling pointers, which can lead to undefined behavior if dereferenced.
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i;
}
free(arr);
int *arr = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i;
}
free(arr);
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i;
}
arr = (int*)malloc(10 * sizeof(int));
int *arr = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = i;
}
Correct Answer: .
Explanation: Option C demonstrates a memory leak because the original memory allocated for `arr` is not freed before reassigning `arr` to a new block of memory. This results in the original memory becoming inaccessible, leading to a memory leak.
int *arr1 = (int*)malloc(5 * sizeof(int));
int *arr2 = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr1[i] = i;
}
memcpy(arr2, arr1, 5 * sizeof(int));
int *arr1 = (int*)calloc(5, sizeof(int));
int *arr2 = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
arr1[i] = i;
}
memcpy(arr2, arr1, 5 * sizeof(int));
int *arr1 = (int*)malloc(5 * sizeof(int));
int *arr2 = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr1[i] = i;
}
arr2 = arr1;
int *arr1 = (int*)calloc(5, sizeof(int));
int *arr2 = (int*)calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
arr1[i] = i;
}
arr2 = arr1;
Correct Answer: .
Explanation: Option A correctly uses `memcpy` to copy the contents from `arr1` to `arr2`. `memcpy` copies a specified number of bytes from the source to the destination, preserving the data.
int *arr = (int*)malloc(3 * sizeof(int));
arr = (int*)realloc(arr, 6 * sizeof(int));
int *arr = (int*)calloc(3, sizeof(int));
arr = (int*)realloc(arr, 6 * sizeof(int));
int *arr = (int*)malloc(3 * sizeof(int));
free(arr);
arr = (int*)realloc(arr, 6 * sizeof(int));
int *arr = (int*)calloc(3, sizeof(int));
free(arr);
arr = (int*)realloc(arr, 6 * sizeof(int));
Correct Answer: .
Explanation: Option A correctly uses `realloc` to resize the memory block to double its original size. The existing data is preserved, and the new elements remain uninitialized.
int *arr = (int*)malloc(5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = 0;
}
int *arr = (int*)malloc(5 * sizeof(int));
int *arr = (int*)calloc(5, sizeof(int));
int *arr = (int*)realloc(NULL, 5 * sizeof(int));
for (int i = 0; i < 5; i++) {
arr[i] = 0;
}
Correct Answer: .
Explanation: Option A correctly allocates memory using `malloc` and then initializes each element to zero. `malloc` does not initialize the memory, so the elements must be explicitly set to zero.
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
arr = (int*)realloc(arr, 5 * sizeof(int));
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
arr = (int*)calloc(5, sizeof(int));
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
arr = (int*)malloc(5 * sizeof(int));
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
free(arr);
arr = (int*)malloc(5 * sizeof(int));
Correct Answer: .
Explanation: Option A correctly resizes the array using `realloc` without losing existing data. `realloc` preserves the existing elements and adds new uninitialized elements to the resized array.
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
free(arr);
arr = NULL;
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
free(arr);
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
free(&arr);
int *arr = (int*)malloc(3 * sizeof(int));
for (int i = 0; i < 3; i++) {
arr[i] = i;
}
Correct Answer: .
Explanation: Option A correctly frees the dynamically allocated memory using `free` and then sets the pointer to `NULL`. Setting the pointer to `NULL` after freeing helps avoid a dangling pointer, which can lead to undefined behavior if dereferenced.
int *arr = (int*)malloc(100000000000 * sizeof(int));
if (arr == NULL) {
printf("Memory allocation failed.\n");
}
int *arr = (int*)malloc(100000000000 * sizeof(int));
int *arr = (int*)malloc(100000000000 * sizeof(int));
free(arr);
int *arr = (int*)malloc(100000000000 * sizeof(int));
if (!arr) {
printf("Memory allocation failed.\n");
}
Correct Answer: .
Explanation: Option D demonstrates the correct way to check if memory allocation using `malloc` fails. If `malloc` returns `NULL`, it indicates that memory allocation failed, and appropriate error handling can be implemented.