printing the address stored in a pointer using write-CodePudding

Can I print an address stored in a pointer using write function only ? (from unistd.h lib).

    int c = 6;
    void *ptr = &c;
    
    printf("%p",ptr);

I'd like to get the same results as the code above using write not printf.

CodePudding user response：

You can simply format the address into a char buffer on your own and pass it to write.

Normally you would just use sprintf to do that but if you don't like printf you might not like sprintf, too.

Nevertheless I will show that first:

  int c = 6;
  void *ptr = &c;

  char outbuf[2*sizeof(void*) 1];
  sprintf("%x", p);
  write(fd, outbuf, strlen(outbuf));

Without sprintf and with MCVE:

#include <stdio.h>
#include <stdint.h>

int main()
{
  int c = 6;
  void *ptr = &c;

  static const char hex_digits[]="0123456789abcdef";
  size_t addr_size = sizeof (void*);
  char outbuf[2*addr_size 1];
  int i;
  uintptr_t val =(intptr_t) ptr;

  int nibble;
  for (i = addr_size - 1; i >= 0; i--)
  {
    nibble = val % 0x10;
    val /= 0x10;
    outbuf[2*i 1] = hex_digits[nibble];

    nibble = val % 0x10;
    val /= 0x10;
    outbuf[2*i] = hex_digits[nibble];
  }
  outbuf[2*addr_size] = 0;
  printf("%s\n", outbuf);
  // replace with write(fd, outbuf, strlen(outbuf))  for your needs
  return 0;
}