How to output long long or int64?

[Richard Bos]

Thanks for all the excellent answers.
My complier is gcc 3.4.2 and can't use another one, so I decide output
like this:
printf("%I64d",c);
Is it all right?

Not in C, it isn't. Maybe in Ganuck, but that's off-topic here; ask in
gnu.gcc.help (If I've remembered the name correctly; readily findable,
anyway).

Richard

 

[Default User]

To Jordan Abel:
It's my mistake.I may wrote "__int64", not "int64"

Looks like you are trying quote messages using Google.

I think you will find the information below of value



Brian

 

[Joe Wright]

Thanks for all the excellent answers.
My complier is gcc 3.4.2 and can't use another one, so I decide output
like this:
printf("%I64d",c);
Is it all right?

Have you tried..

#include <stdio.h>
long long a;
int main(void)
{
a = 999999999;
a *= 5;
printf("%lld\n", a);
return 0;
}

...a real C program?

With GCC 3.1 (DJGPP) this prints..
4999999995

 

[Betaver]

Joe Wright wrote:

With GCC 3.1 (DJGPP) this prints..
4999999995

Maybe the problem is my header file. But "%lld" isn't supported here.
Certainly "%I64d" works well.

 

[Betaver]

With GCC 3.1 (DJGPP) this prints..
4999999995

With Dev-C++ 4.9.9.2 Gcc 3.4.2

 

[jacob navia]

With Dev-C++ 4.9.9.2 Gcc 3.4.2

The printf you are calling is Microsoft printf in CRTDLL.DLL
Apparently you are using the mingw version of gcc without glibc.
Use I64 in that system since it works.

 

[jacob navia]

<SEMI-OT>
printf is part of the runtime library; it's not part of gcc.
</SEMI-OT>

Under windows they are using crtdll.dll by Microsoft.