How to print space in uint64_t (C)

Question

When I try this:

#include <stdio.h>
#include <inttypes.h>

int main (int argc, char **argv)
{
  uint64_t key = 0xf0000000ffffffff;

  printf ("key=%" PRIx64, key, "\t\n  %" PRIx64, "\n", key);
}

It prints key=f0000000ffffffff; all whitespaces are removed. How can I print them? I am using gcc (Ubuntu/Linaro 4.7.3-1ubuntu1) 4.7.3.

Never ignore the compiler warnings ... such as `too many arguments for format [-Wformat-extra-args]` for a decent gcc — Ferenc Deak, Jul 29 '14 at 06:56
@fritzone `gcc`. I stated it in my question. My OS is `Mint Olivia Xfce (64 bit)`. — hola, Jul 29 '14 at 06:58
yup, saw it too late :) Removed the comment. gcc 4.7.3 is actually a pretty decent compiler, it should have given a warning ... — Ferenc Deak, Jul 29 '14 at 06:59

score 4 · Accepted Answer · answered Jul 29 '14 at 06:51

The format string you provide to printf is only

"key=%" PRIx64

All the rest are arguments for the format string, ans since the format string only contain one conversion specifier the first argument is printed and the rest are discarded.

I think you want e.g.

printf("key=%" PRIx64 "\t\n  %" PRIx64 "\n", key, key);

Note the lack of commas in the format string.

How to print space in uint64_t (C)

1 Answers1