[c] How to print a int64_t type in C

C99 standard has integer types with bytes size like int64_t. I am using the following code:

#include <stdio.h>
#include <stdint.h>
int64_t my_int = 999999999999999999;
printf("This is my_int: %I64d\n", my_int);

and I get this compiler warning:

warning: format ‘%I64d’ expects type ‘int’, but argument 2 has type ‘int64_t’

I tried with:

printf("This is my_int: %lld\n", my_int); // long long decimal

But I get the same warning. I am using this compiler:

~/dev/c$ cc -v
Using built-in specs.
Target: i686-apple-darwin10
Configured with: /var/tmp/gcc/gcc-5664~89/src/configure --disable-checking --enable-werror --prefix=/usr --mandir=/share/man --enable-languages=c,objc,c++,obj-c++ --program-transform-name=/^[cg][^.-]*$/s/$/-4.2/ --with-slibdir=/usr/lib --build=i686-apple-darwin10 --program-prefix=i686-apple-darwin10- --host=x86_64-apple-darwin10 --target=i686-apple-darwin10 --with-gxx-include-dir=/include/c++/4.2.1
Thread model: posix
gcc version 4.2.1 (Apple Inc. build 5664)

Which format should I use to print my_int variable without having a warning?

In windows environment, use

%I64d

in Linux, use

%lld

The C99 way is

#include <inttypes.h>
int64_t my_int = 999999999999999999;
printf("%" PRId64 "\n", my_int);

Or you could cast!

printf("%ld", (long)my_int);
printf("%lld", (long long)my_int); /* C89 didn't define `long long` */
printf("%f", (double)my_int);

If you're stuck with a C89 implementation (notably Visual Studio) you can perhaps use an open source <inttypes.h> (and <stdint.h>): http://code.google.com/p/msinttypes/

With C99 the %j length modifier can also be used with the printf family of functions to print values of type int64_t and uint64_t:

#include <stdio.h>
#include <stdint.h>

int main(int argc, char *argv[])
{
    int64_t  a = 1LL << 63;
    uint64_t b = 1ULL << 63;

    printf("a=%jd (0x%jx)\n", a, a);
    printf("b=%ju (0x%jx)\n", b, b);

    return 0;
}

Compiling this code with gcc -Wall -pedantic -std=c99 produces no warnings, and the program prints the expected output:

a=-9223372036854775808 (0x8000000000000000)
b=9223372036854775808 (0x8000000000000000)

This is according to printf(3) on my Linux system (the man page specifically says that j is used to indicate a conversion to an intmax_t or uintmax_t; in my stdint.h, both int64_t and intmax_t are typedef'd in exactly the same way, and similarly for uint64_t). I'm not sure if this is perfectly portable to other systems.

//VC6.0 (386 & better)

    __int64 my_qw_var = 0x1234567890abcdef;

    __int32 v_dw_h;
    __int32 v_dw_l;

    __asm
        {
            mov eax,[dword ptr my_qw_var + 4]   //dwh
            mov [dword ptr v_dw_h],eax

            mov eax,[dword ptr my_qw_var]   //dwl
            mov [dword ptr v_dw_l],eax

        }
        //Oops 0.8 format
    printf("val = 0x%0.8x%0.8x\n", (__int32)v_dw_h, (__int32)v_dw_l);

Regards.

Coming from the embedded world, where even uclibc is not always available, and code like

uint64_t myval = 0xdeadfacedeadbeef; printf("%llx", myval);

is printing you crap or not working at all -- i always use a tiny helper, that allows me to dump properly uint64_t hex:

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>

char* ullx(uint64_t val)
{
    static char buf[34] = { [0 ... 33] = 0 };
    char* out = &buf[33];
    uint64_t hval = val;
    unsigned int hbase = 16;

    do {
        *out = "0123456789abcdef"[hval % hbase];
        --out;
        hval /= hbase;
    } while(hval);

    *out-- = 'x', *out = '0';

    return out;
}