This program will finally allocate the memory of ___ bytes that “ptr” will point to. #include #include int main() { char *ptr; ptr = (char*)malloc(sizeof(char)*4); ptr = (char*)realloc(ptr,sizeof(char)*8); return 0; }

Question

BlackTom AI · Accepted Answer

sizeof(char) is defined as 1 byte in C, so any expression sizeof(char) * N equals N bytes. Therefore sizeof(char)*8 = 1 * 8 = 8 bytes.

1) Relevant knowledge points and concepts
- sizeof(char) is 1 (C standard guarantees this).
- malloc(n) allocates n bytes; realloc(ptr, n) resizes the allocation to n bytes (possibly moving it).
- The final size of the allocation equals the size argument passed to the last successful allocation/reallocation call.

2) Step-by-step reasoning
- The code first calls malloc(sizeof(char)*4). Since sizeof(char) = 1, this is malloc(1 * 4) = malloc(4), allocating 4 bytes initially.
- Next the code calls realloc(ptr, sizeof(char)*8). That is realloc(ptr, 1 * 8) = realloc(ptr, 8), requesting an 8-byte block.
- On success, realloc returns a pointer to memory of the requested size (8 bytes). The variable ptr is assigned this pointer, so after the realloc the memory that ptr points to is 8 bytes in size.

3) Calculations (intermediate steps)
- sizeof(char) = 1
- sizeof(char) * 4 = 1 * 4 = 4 bytes (initial malloc)
- sizeof(char) * 8 = 1 * 8 = 8 bytes (realloc request and final size)

4) Why the correct answer is correct
Because the final allocation size is determined by the last call realloc(ptr, sizeof(char)*8), which requests 8 bytes. realloc returns memory sized exactly as requested (on success), so ptr points to 8 bytes.

5) Why the other options are incorrect
- 16: No step requests 16 bytes; likely a confusion with 2-byte chars or doubling twice, but sizeof(char) is 1.
- 4: This was the initial allocation size, but it was resized by realloc to 8 bytes, so 4 is not the final size.
- 32: No operation requests anywhere near 32 bytes; incorrect.

6) Conclusion
The final memory allocated for ptr is 8 bytes, matching the provided answer: 8.

This program will finally allocate the memory of ___ bytes that “ptr” will point to. #include<stdio.h> #include<stdlib.h> int main() { char ptr; ptr = (char)malloc(sizeof(char)4); ptr = (char)realloc(ptr,sizeof(char)*8); return 0; }单项选择题

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This program will finally allocate the memory of ___ bytes that “ptr” will point to. #include<stdio.h> #include<stdlib.h> int main() { char *ptr; ptr = (char*)malloc(sizeof(char)*4); ptr = (char*)realloc(ptr,sizeof(char)*8); return 0; }单项选择题