Re: [Mingw-w64-public] 64-bit printf question...

[Kai T. <kti...@go...>]

Hello David,

2009/10/25 David Cleaver <wr...@mo...>:
> Hello,
>
> I'm having problems printing out values that are larger than 32 bits.  My usual
> printf using %llu only outputs the lower 32 bits of my 64-bit numbers.
>
> Here's an example program that produces incorrect output.
>
> #include <stdio.h>
>
> typedef unsigned long long u64_t;
>
> int main(int argc, char* argv[])
> {
>   u64_t y=6981463658333LL;
>   printf("y=%llu\n", y);
>   printf("y=%PRIu64\n", y);
>   printf("y=%qd\n", y);
>   printf("y=%x\n", y);
>   return 0;
> }
>
> Here is the output I get.
> y=2141802333
> y=PRIu64
> y=qd
> y=7fa94f5d
>
> Can anyone see what I'm doing wrong?  What is the proper way to output 64-bit
> numbers?  I'm using mingw-w64-bin_i686-mingw_20091025.zip for my development
> environment and I've called the compiler like:
> x86_64-w64-mingw32-gcc.exe -o test.exe test.c
>
> Thanks for any help.
>
> -David C.

The reason for this is the msvcrt you are using. The %ll width
specifier is support by MS until msvcrt based on msvcr80.dll. The
common width specifier, which works also for older msvcrt.dll
versions, is %I64.
If you want to use POSIX like printf formatters, like "%ll" or "%Lg",
then you should use our POSIX emulation version for it. You need just
to define by command line, or before including the first header file,
the macro __USE_MINGW_ANSI_STDIO with value one.

For the example you gave it would look like that

#define __USE_MINGW_ANSI_STDIO 1
#include <stdio.h>

typedef unsigned long long u64_t;

int main(int argc, char* argv[])
{
  u64_t y=6981463658333LL;
  printf("y=%llu\n", y);
  printf("y=%PRIu64\n", y);
  printf("y=%qd\n", y);
  printf("y=%x\n", y);
  return 0;
}

I hope I could help you by this.

Cheers,
Kai

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