Reputation: 7143
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?
Upvotes: 9
Views: 22305
Reputation: 28239
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
Upvotes: 21
Reputation: 117641
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);
Upvotes: 0
Reputation: 441
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.
Upvotes: 0
Reputation: 137272
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.
Upvotes: 0