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

There's another way using arrays and pointers:

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

int main(void) {
    // Two uint32_t to one uint64_t
    uint32_t val1[2] = {1000, 90000};
    uint64_t *val1_u64_ptr = (uint64_t*)val1; //intermediate pointer cast to avoid Wstrict-aliasing warnings
    uint64_t val2 = *val1_u64_ptr;
    printf("val2: %" PRIu64 "\n", val2);
    // val2: 386547056641000


    // back to uint32_t array from uint64_t
    uint64_t val3 = 386547056641000ull;
    uint32_t *val4 = (uint32_t*)&val3;
    printf("val4: %" PRIu32 ", %" PRIu32 "\n", val4[0], val4[1]);
    // val4: 1000, 90000
    
    return 0;
}

This code for me is much easier to understand and read. You are just creating a contiguous space in memory with two 32-bit unsigned int and then this same memory space is read as a single 64-bit unsigned int value and vice-versa. There are no operations involved only memory being read as different types.

EDIT

Forgot to mention that this is great if you already have a 64-bit array read from somewhere then you could easily read everything as 32-bit array pairs:

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

int main() {
    uint64_t array64[] = {
        386547056641000ull,
        93929935171414ull,
        186655006591110ull,
        73141496240875ull,
        161460097995400ull,
        351282298325439ull,
        97310615654411ull,
        104561732955680ull,
        383587691986172ull,
        386547056641000ull
    };

    int n_items = sizeof(array64) / sizeof(array64[0]);
    
    uint32_t* array32 = (uint32_t*)&array64;
    for (int ii = 0; ii < n_items * 2; ii += 2) {
        printf("[%" PRIu32 ", %" PRIu32 "]\n", array32[ii], array32[ii + 1]);
    }

    return 0;
}

Output:

[1000, 90000]
[3295375190, 21869]
[22874246, 43459]
[2498157291, 17029]
[3687404168, 37592]
[1218152895, 81789]
[3836596235, 22656]
[754134560, 24345]
[4162780412, 89310]
[1000, 90000]

Using union struct

Still better and more readable would be to use a struct union as from https://stackoverflow.com/a/2810339/2548351:

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

typedef union {
  int64_t big;
  struct {
    int32_t x;
    int32_t y;
  };
} xy_t;

int main() {
    // initialize from 64-bit
    xy_t value = {386547056641000ull};
    printf("[%" PRIu32 ",%" PRIu32 "]\n", value.x, value.y);
    // [1000, 90000]
    
    // initialize as two 32-bit
    xy_t value2 = {.x = 1000, .y = 90000};
    printf("%" PRIu64, value.big);
    // 386547056641000

    return 0;
}

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

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.

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