请问C轮浮点常量 [英] Does C round floating-point constants

问题描述

<一个href=\"http://stackoverflow.com/questions/22337418/golang-floating-point-$p$pcision-float32-vs-float64\">A关于围棋浮点precision 让我不知道Ç如何处理这个问题的问题。

A question about floating-point precision in Go made me wonder how C handles the problem.

在C以下code：

float a = 0.1;

威尔 A 有最亲密的IEEE 754二进制重新presentation：

Will a have the closest IEEE 754 binary representation of:

00111101110011001100110011001101 (Decimal:  0.10000000149011612)

还是会只是它作物：

or will it just crop it to:

00111101110011001100110011001100 (Decimal:  0.09999999403953552)

还是会依赖于编译器/平台有什么不同？

Or will it differ depending on compiler/platform?

推荐答案

这是实现允许做任何（或甚至一个更多关闭）：

An implementation is allowed to do either (or even be off by one more):

有关十进制浮点常量，也为十六进制浮点常量时FLT_RADIX不是2的幂，其结果是要么最近重新presentable值，或更大或更小重presentable紧邻值最近重新presentable价值，在实现定义的方式选择。

For decimal floating constants, and also for hexadecimal floating constants when FLT_RADIX is not a power of 2, the result is either the nearest representable value, or the larger or smaller representable value immediately adjacent to the nearest representable value, chosen in an implementation-defined manner.

（C11，与教派; 6.4.4.2/3）

(C11, §6.4.4.2/3)

C99以来，我们已经有十六进制浮点常量，使您可以指定precisely你想（假设实现提供二进制浮点:)）位，所以你可以说，例如：

Since C99, we've had hexadecimal floating point constants so that you can specify precisely the bits you want (assuming that the implementation offers binary floating point :) ), so you could have said, for example:

float a = 0x1.99999Ap-4;

有关IEEE 754 32位彩车：

For IEEE 754 32-bit floats:

#include <stdio.h>
int main() {
  float a = 0.1;
  float low  = 0x0.1999999p0;
  float best = 0x0.199999ap0;
  float high = 0x0.199999bp0;
  printf("a is %.6a or %.16f, which is either %.16f, %.16f or %.16f\n",
          a, a, low, best, high);
  return 0;
}

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