如何使用gcc __uint128_t号打印？ [英] how to print __uint128_t number using gcc?

问题描述

有 PRIu128 ，其行为类似于从＆LT到 PRIu64 ; inttypes.h＆GT; ：

Is there PRIu128 that behaves similar to PRIu64 from <inttypes.h>:

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

或数字转换手动位数：

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);
}

是唯一的选择？

注意 uint128_t 是我自己的类型定义 __ uint128_t 。

Note that uint128_t is my own typedef for __uint128_t.

推荐答案

没有有没有在库支持打印这些类型。在C标准的意义上，它们甚至没有扩展整型。

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.

您启动从背面印刷的想法是好的，但你可以使用更大的块。在某些测试中的P99我有一个使用

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);

作为适合10最大功率的 uint64_t中。

由于小数，这些数字大得无法阅读很快使另一个更简单，选择是打印出来的十六进制。然后，你可以做这样的事情。

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);

获得下半部，然后基本上是相同的用

to get the lower half and then basically the same with

  uint64_t high = (x >> 64);

有上半部分

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