Getting the warning “cast to pointer from integer of different size” from the following code

问题

The code is:

           Push(size, (POINTER)(GetCar(i) == term_Null()? 0 : 1));

Here is the C code push returns ABC which is

 typedef POINTER  *ABC
 typedef void * POINTER
 ABC size;
 Push(ABC,POINTER);
 XYZ GetCar(int);
 typedef struct xyz *XYZ;
 XYZ term_Null(); 
 long int i;

What is the reason for the particular warning?

回答1:

You can use intptr_t to ensure the integer has the same width as pointer. This way, you don't need to discover stuff about your specific platform, and it will work on another platform too (unlike the unsigned long solution).

#include <stdint.h>

Push(size, (POINTER)(intptr_t)(GetCar(i) == term_Null()? 0 : 1));

Taken from the C99 Standard:

7.18.1.4 Integer types capable of holding object pointers

1 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 pointer to void, and the result will compare equal to the original pointer:

intptr_t

回答2:

You are trying to cast an integer value (0 or 1) to a void pointer.

This expression is always an int with value 0 or 1: (GetCar(i) == term_Null()? 0 : 1)

And you try casting it to void pointer (POINTER) (typedef void * POINTER).

Which is illegal.

回答3:

Since this question uses the same typedefs as your 32bit to 64bit porting question I assume that you're using 64 bit pointers. As MByd wrote you're casting an int to a pointer and since int isn't 64 bit you get that particular warning.

回答4:

What are you trying to do? Pointers are not integers, and you are trying to make a pointer out of 0 or 1, depending on the situation. That is illegal.

If you were trying to pass a pointer to a ABC containing 0 or 1, use this:

ABC tmp = GetCar(i) == term_Null()? 0 : 1;
Push(size, &tmp);

来源：https://stackoverflow.com/questions/5701450/getting-the-warning-cast-to-pointer-from-integer-of-different-size-from-the-fo

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gcc

gcc-warning