How to print a int64_t type in C

For int64_t type:

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

for uint64_t type:

#include <inttypes.h>uint64_t t;printf("%" PRIu64 "\n", t);

you can also use PRIx64 to print in hexadecimal.

cppreference.com has a full listing of available macros for all types including intptr_t (PRIxPTR). There are separate macros for scanf, like SCNd64.

A typical definition of PRIu16 would be "hu", so implicit string-constant concatenation happens at compile time.

For your code to be fully portable, you must use PRId32 and so on for printing int32_t, and "%d" or similar for printing int.

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.