How to combine two 32-bit integers into one 64-bit integer?

Question

I have a count register, which is made up of two 32-bit unsigned integers, one for the higher 32 bits of the value (most significant word), and other for the lower 32 bits of the value (least significant word).

What is the best way in C to combine these two 32-bit unsigned integers and then display as a large number?

In specific:

leastSignificantWord = 4294967295; //2^32-1

printf("Counter: %u%u", mostSignificantWord,leastSignificantWord);

This would print fine.

When the number is incremented to 4294967296, I have it so the leastSignificantWord wipes to 0, and mostSignificantWord (0 initially) is now 1. The whole counter should now read 4294967296, but right now it just reads 10, because I'm just concatenating 1 from mostSignificantWord and 0 from leastSignificantWord.

How should I make it display 4294967296 instead of 10?

Answer 1

long long val = (long long) mostSignificantWord << 32 | leastSignificantWord;
printf( "%lli", val );

Answer 2

my take:

unsigned int low = <SOME-32-BIT-CONSTRANT>
unsigned int high = <SOME-32-BIT-CONSTANT>

unsigned long long data64;

data64 = (unsigned long long) high << 32 | low;

printf ("%llx\n", data64); /* hexadecimal output */
printf ("%lld\n", data64); /* decimal output */

Another approach:

unsigned int low = <SOME-32-BIT-CONSTRANT>
unsigned int high = <SOME-32-BIT-CONSTANT>

unsigned long long data64;
unsigned char * ptr = (unsigned char *) &data;

memcpy (ptr+0, &low, 4);
memcpy (ptr+4, &high, 4);

printf ("%llx\n", data64); /* hexadecimal output */
printf ("%lld\n", data64); /* decimal output */

Both versions work, and they will have similar performance (the compiler will optimize the memcpy away).

The second version does not work with big-endian targets but otoh it takes the guess-work away if the constant 32 should be 32 or 32ull. Something I'm never sure when I see shifts with constants greater than 31.

Answer 3

Instead of attempting to print decimal, I often print in hex.

Thus ...

printf ("0x%x%08x\n", upper32, lower32);

Alternatively, depending upon the architecture, platform and compiler, sometimes you can get away with something like ...

printf ("%lld\n", lower32, upper32);

or

printf ("%lld\n", upper32, lower32);

However, this alternative method is very machine dependent (endian-ness, as well as 64 vs 32 bit, ...) and in general is not recommended.

Hope this helps.

Answer 4

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;
}

Output

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