How to represent address of a variable as a String?-CodePudding

I have to reproduce the printf function in C. I am getting problem with printing addresses (%p). So i have an address of variable and I need to return it as an array of char. Is it even possible?

char *ptrToString(void * x)
{   
    char *t = (char *)x;
    printf("Here: %s\n",t);
    return "abc";
}

char *getPtr(unsigned long long x)
{
    return ptrToString(&x);
}

CodePudding user response：

Pointers are numeric types, so in printf-like functions use either %p (preferred), %x or %d to print them.

Your ptrToString could look something like this:


#include <stdio.h>
#include <stdlib.h>

const char *x = "test";

char *ptrToString(void *x) {
    char *s = malloc(sizeof(void*) * sizeof(char));
    sprintf(s, "%p", x);
    return s;
}

int main() {
    char *a = ptrToString( (void*) x);

    if (!a)
        return 1;

    printf("%s\n", a);
    free(a);
    return 0;
}

CodePudding user response：

First a small note:

char *getPtr(unsigned long long x)
{
    return ptrToString(&x);
}

int main() 
{
   long long someLong = 1;
   char *ptr = getPtr(someLong);
   return 0;
}

Here ptrToString will return the address of variable x on the stack rather than the address of variable someLong because the value of the variable will get copied on the stack as a totally separate entity.

As for the answer itself, an address is actually just a number representing byte offsets in memory. To convert this number to a string, you need a variant integer to ascii-like function where you convert to base 16 instead of base 10.

I assume sprintf is not an option, so you'll have to implement your own variant of a decimal to hex string converter. Cast any address to a wide unsigned integer type and convert that number to hex.