Ç - 舍入整数除法（而不是截断） [英] C - Rounding integer division (instead of truncating)

问题描述

我很好奇，想知道我可以圆一个数字，最近的第十整数。举例来说，如果我有：

  int类型的= 59/4;
 

这将是14.75浮点计算;我怎么能存储的号码15在一？

解决方案

  int类型的= 59.0f / 4.0F + 0.5F;
 

这只是分配给一个int的时候，因为它放弃了后面的所有作品'。

编辑：
这种解决方案将只在例的最简单的工作。一个更强大的解决方案是：

  unsigned int类型round_div（unsigned int类型红利，无符号整型除数）
{
    收益率（红利+（除数/ 2））/除数;
}
 

I was curious to know how I can round a number to the nearest tenth whole number. For instance, if I had:

int a = 59 / 4;

which would be 14.75 calculated in floating point; how can I store the number as 15 in "a"?

解决方案

int a = 59.0f / 4.0f + 0.5f;

This only works when assigning to an int as it discards anything after the '.'

Edit: This solution will only work in the simplest of cases. A more robust solution would be:

unsigned int round_div(unsigned int dividend, unsigned int divisor)
{
    return (dividend + (divisor / 2)) / divisor;
}

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