How to printf(“%0x”) with the number of digits determined from the type?

Question

I can specify the number of digits like printf("%08x", ...) . But I'd like to automatically adjust the number of digits according to the data type, (eg, long (64 bits) should automatically have 64/4=16 digits). Is there a way to do so? Thanks.

Answer 1

How to printf(“%0x”) with the number of digits determined from the type?

1) Find the bit width

Use sizeof, CHAR_BIT to find the bit width sizeof(x)*CHAR_BIT ¹ or see below alternative.

2) Find the nibble width

(bit_width + 3)/4 for the nibble width - number of hexadecimal characters.

3) Pass the width

Use "*" to pass in the width and print using the widest type, uintmax_t .

#include <float.h>
#include <limits.h>
#include <stdio.h>

#define PRINTF_XW(x) printf("%0*jx\n", (int)(sizeof(x)*CHAR_BIT+3)/4, (uintmax_t) (x))

int main(void) {
  unsigned char uc = 1;
  unsigned short us = 2;
  unsigned u = 3;
  unsigned long ul = 4;
  unsigned long long ull = 5;
  uintmax_t uj = 6;

  PRINTF_XW(uc);
  PRINTF_XW(us);
  PRINTF_XW(u);
  PRINTF_XW(ul);
  PRINTF_XW(ull);
  PRINTF_XW(uj);
  return 0;
}

Sample output, may differ amongst platforms.

01
0002
00000003
0000000000000004
0000000000000005
0000000000000006

---

Alternative

Another approach would use _Generic - tis a bit more work - and more rewards.
See unsigned long long uf:42; example below.

#include <float.h>
#include <limits.h>
#include <stdint.h>

int hexwidth(uintmax_t u) {
  int count = 3;
  while (u) {
    u >>= 1;
    count++;
  }
  return count/4;
}

// Default case useful for other unsigned types wider than `unsigned`
#define HEXWIDTH(s) _Generic((s), \
    unsigned char: hexwidth(UCHAR_MAX), \
    unsigned short: hexwidth(USHRT_MAX), \
    unsigned : hexwidth(UINT_MAX), \
    unsigned long: hexwidth(ULONG_MAX), \
    unsigned long long: hexwidth(ULLONG_MAX), \
    default: hexwidth((s)*0u - 1u) \
    )

#define PRINTF_XW(x) printf("%0*jx\n", HEXWIDTH(x), (uintmax_t) (x))

With language extensions that allow wider than unsigned fields - works nicely too.

int main(void) {
  struct {
      unsigned long long uf:42;
  } s1 = {7};
  struct {
      unsigned long long uf:51;
  } s2 = {8};

  PRINTF_XW(s1.uf);
  PRINTF_XW(s2.uf);
}

Output

00000000007   <-- 11 digits
0000000000008 <-- 13 digits

Macro-magic could replace hexwidth(uintmax_t u) as follows with a compile time calculation:

#define NW3(x) (((x) > 0xFu) ? 2 : 1)
#define NW4(x) (((x) > 0xFFu) ? NW3((x)>>8)+2 : NW3(x))
#define NW5(x) (((x) > 0xFFFFu) ? NW4((x)>>16)+4 : NW4(x))
#define NW6(x) (((x) > 0xFFFFFFFFu) ? NW5((x)>>32)+8 : NW5(x))
#define NW(x) (((x) > 0xFFFFFFFFFFFFFFFFu) ? NW6((((x)+0llu)>>32)>>32)+16 : NW6((x)+0llu))

// Default case useful for all unsigned types as wide/wider than `unsigned`
#define HEXWIDTH(s) _Generic((s), \
    unsigned char: NW(UCHAR_MAX), \
    unsigned short: NW(USHRT_MAX), \
    default: NW((s)*0u - 1u) \
    )

---

¹ This works well when there is no padding in the type - common amongst standard unsigned integer types.

Answer 2

You should probably look into C's <inttypes.h>

#include <inttypes.h>

unit8_t value = 0x34;
printf("The value formatted for 8 bits is %" PRIx8 ".", value);

Note the PRIx8 is a string, which correctly will format for "printf" "hexadecimal" "eight bits".

Note that the lack of commas between the strings are not a mistake. This is a formatting technique that uses a feature in C called "automatic string concatenation", such that the three strings will get slapped together. The PRIx8 is just a convenience which means the correct formatting for an 8 bit hexadecimal value.

To get the similar output formatted for 32 bits, you would use PRIx32 .

For a brief overview of the inttypes.h output formats, you can look at Good introduction to <inttypes.h>