是否有一个函数来圆在C浮动或我需要写我自己？ [英] Is there a function to round a float in C or do I need to write my own?

问题描述

有圆在C浮动或我需要写我自己的函数吗？

Is there a function to round a float in C or do I need to write my own?

浮动CONVER = 45. 59 2346543;

float conver = 45.592346543;

我想实际值四舍五入至小数点后一位，CONVER = 45. 6

I would like to round the actual value to one decimal place, conver = 45.6.

推荐答案

作为罗布提到的，你可能只是想的打印的浮子1位小数。在这种情况下，你可以做类似如下：

As Rob mentioned, you probably just want to print the float to 1 decimal place. In this case, you can do something like the following:

#include <stdio.h>
#include <stdlib.h>

int main()
{
  float conver = 45.592346543;
  printf("conver is %0.1f\n",conver);
  return 0;
}

如果您想实际一轮存储的值，这是一个稍微复杂一点。首先，你的一位小数就地重新presentation很少有浮点精确的模拟。如果你只是想获得尽可能接近，这样的事情可能做的伎俩：

If you want to actually round the stored value, that's a little more complicated. For one, your one-decimal-place representation will rarely have an exact analog in floating-point. If you just want to get as close as possible, something like this might do the trick:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
  float conver = 45.592346543;
  printf("conver is %0.1f\n",conver);

  conver = conver*10.0f;
  conver = (conver > (floor(conver)+0.5f)) ? ceil(conver) : floor(conver);
  conver = conver/10.0f;

  //If you're using C99 or better, rather than ANSI C/C89/C90, the following will also work.
  //conver = roundf(conver*10.0f)/10.0f;

  printf("conver is now %f\n",conver);
  return 0;
}

我怀疑这第二个例子是你在找什么，但我把它的完整性。如果您确实需要重新$ P $这样psenting你的号码在内部，而不是仅仅输出，可以考虑使用定点再presentation 代替。

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