Casting a pointer to an int

问题

I am writing my own functions for malloc and free in C for an assignment. I need to take advantage of the C sbrk() wrapper function. From what I understand sbrk() increments the program's data space by the number of bytes passed as an argument and points to the location of the program break.

If I have the following code snippet:

#define BLOCK_SIZE 20

int x;

x = (int)sbrk(BLOCK_SIZE + 4);

I get the compiler error warning: cast from pointer to integer of different size. Why is this and is there anyway I can cast the address pointed to by sbrk() to an int?

回答1:

I get the compiler error warning: cast from pointer to integer of different size.
Why is this

Because pointer and int may have different length, for example, on 64-bit system, sizeof(void *) (i.e. length of pointer) usually is 8, but sizeof(int) usually is 4. In this case, if you cast a pointer to an int and cast it back, you will get a invalid pointer instead of the original pointer.

and is there anyway I can cast the address pointed to by sbrk() to an int?

If you really need to cast a pointer to an integer, you should cast it to an intptr_t or uintptr_t, from <stdint.h>.

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From <stdint.h>(P):

• Integer types capable of holding object pointers

The following type designates a signed integer type with the property that any valid pointer to void can be converted to this type, then converted back to a pointer to void, and the result will compare equal to the original pointer: intptr_t

The following type designates an unsigned integer type with the property that any valid pointer to void can be converted to this type, then converted back to a pointer to void, and the result will compare equal to the original pointer: uintptr_t

On XSI-conformant systems, the intptr_t and uintptr_t types are required; otherwise, they are optional.

来源：https://stackoverflow.com/questions/22624737/casting-a-pointer-to-an-int