J.F. Sebastian - 11 months ago 218

C Question

Is there

`PRIu128`

`PRIu64`

`<inttypes.h>`

`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

`uint128_t`

`__uint128_t`

Answer

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.

Source (Stackoverflow)