J.F. Sebastian J.F. Sebastian - 1 year ago 381
C Question

how to print __uint128_t number using gcc?

Is there

that behaves similar to

printf("%" PRIu64 "\n", some_uint64_value);

Or converting manually digit by digit:

int print_uint128(uint128_t n) {
if (n == 0) return printf("0\n");

char str[40] = {0}; // log10(1 << 128) + '\0'
char *s = str + sizeof(str) - 1; // start at the end
while (n != 0) {
if (s == str) return -1; // never happens

*--s = "0123456789"[n % 10]; // save last digit
n /= 10; // drop it
return printf("%s\n", s);

is the only option?

Note that
is my own typedef for

Answer Source

No there isn't support in the library for printing these types. They aren't even extended integer types in the sense of the C standard.

Your idea for starting the printing from the back is a good one, but you could use much larger chunks. In some tests for P99 I have such a function that uses

uint64_t const d19 = UINT64_C(10000000000000000000);

as the largest power of 10 that fits into an uint64_t.

As decimal, these big numbers get unreadable very soon so another, easier, option is to print them in hex. Then you can do something like

  uint64_t low = (uint64_t)x;
  // This is UINT64_MAX, the largest number in 64 bit
  // so the longest string that the lower half can occupy
  char buf[] = { "18446744073709551615" };
  sprintf(buf, "%" PRIX64, low);

to get the lower half and then basically the same with

  uint64_t high = (x >> 64);

for the upper half.

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