C getting type of a pointer

Question

I want to implement a function that takes either int or a char pointer as the input and stores then as a void pointer but I also want to store the type of the pointer that has been passed so is there a way that I could find the type of the pointer

Answer 1

You can perform sort-of-overloading on types using _Generic macro if you are using a C11 compliant compiler, but the type has to be stored separately, be it a void* or an union with some sort of tag.

The closest to what you want is sth like this:

#include <stdio.h>
#include <assert.h>

struct TaggedUnion
{
    void* contents;
    /* union {int* i;char* c;} contents; */

    enum Type {CHAR_TYPE, INT_TYPE} type;
};

struct TaggedUnion storeI(int* x)
{
    struct TaggedUnion ret = {.contents =x, .type=INT_TYPE};
    return ret;
}

struct TaggedUnion storeC(char* x)
{
    struct TaggedUnion ret = {.contents =x, .type=CHAR_TYPE};
    return ret;
}

#define storeTagged(x) _Generic((x),\
    int*: storeI,\
    char*: storeC)(x)

int* getInt(const struct TaggedUnion x)
{
    return x.type == INT_TYPE ? x.contents : NULL;
} 

char* getChar(const struct TaggedUnion x)
{
    return x.type == CHAR_TYPE ? x.contents : NULL;
} 

void fi(int* arg)
{
    printf("i\n");
}

void fc(char* arg)
{
    printf("c\n");
}


#define ff(x) _Generic((x),\
    int*: fi,\
    char*: fc)(x)

int main(int argc, const char* argv[])
{
    printf("entry\n");
    int* i;
    char* c;

    ff(i);
    ff(c);

    struct TaggedUnion ti = storeTagged(i);
    struct TaggedUnion tc = storeTagged(c);

    assert(ti.type == INT_TYPE);
    assert(tc.type == CHAR_TYPE);

    return 0;
}

https://godbolt.org/z/5dcrc84fo