C Programming – Why One printf() Can’t Print Two 64-bit Values Simultaneously?

It might be advantageous to use unsigned integers with explicit sizes in this case:

#include <stdio.h>
#include <inttypes.h>

int main(void) {
  uint32_t leastSignificantWord = 0;
  uint32_t mostSignificantWord = 1;
  uint64_t i = (uint64_t) mostSignificantWord << 32 | leastSignificantWord;
  printf("%" PRIu64 "\n", i);

  return 0;
}

4294967296

Break down of (uint64_t) mostSignificantWord << 32 | leastSignificantWord

• (typename) does typecasting in C. It changes value data type to typename.

  (uint64_t) 0x00000001 -> 0x0000000000000001

• << does left shift. In C left shift on unsigned integers performs logical shift.

  0x0000000000000001 << 32 -> 0x0000000100000000

• | does 'bitwise or' (logical OR on bits of the operands).

  0b0101 | 0b1001 -> 0b1101

There are a couple of approaches.

You could write your code in C99-conforming fashion, and then supply system-specific hacks when the compiler-writers let you down. (Sadly, that's rather common in C99.)

#include <stdint.h>
#include <inttypes.h>

printf("My value is %10" PRId64 "\n", some_64_bit_expression);

If one of your target systems has neglected to implement <inttypes.h> or has in some other way fiendishly slacked off because some of the type features are optional, then you just need a system-specific #define for PRId64 (or whatever) on that system.

The other approach is to pick something that's currently always implemented as 64-bits and is supported by printf, and then cast. Not perfect but it will often do:

printf("My value is %10lld\n", (long long)some_64_bit_expression);