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

[Kai Tietz]

Hello David,

2009/10/25 David Cleaver wrai...@morpheus.net:
 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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Re: [Mingw-w64-public] 64-bit printf question...

[Ozkan Sezer]

On Sun, Oct 25, 2009 at 5:55 PM, David Cleaver wrai...@morpheus.net wrote:
 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);

%llu is not supported by printf in msvcrt.dll.  You can use
%I64u which is the one supported by ms. Or you can use
__mingw_printf(), instead, either explicitly, or indirectly by
defining __USE_MINGW_ANSI_STDIO as 1, for example
by adding -D__USE_MINGW_ANSI_STDIO=1 to your
CFLAGS, which will automaticly replace all printf family
of functions with their __mingw_ counterparts.


   printf(y=%PRIu64\n, y);

Wrong syntax.  First you must #include inttypes.h for
this. Then, the correct form should be like

printf(y=% PRIu64 \n, y);

   printf(y=%qd\n, y);

Same as %ll stuff.

   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.

Use -Wall (or -Wformat) to get warnings about what is
supported and what is not.  Hope these help.

--
Ozkan

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