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

问题描述

我很想知道如何将数字舍入到最接近的第十整数。例如，如果我有：

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;

以浮点计算的14.75;如何将数字存储为a中的15？

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

推荐答案

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

这仅适用于分配给int，因为它会丢弃'。'之后的任何内容。

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

修改：
此解决方案仅适用于最简单的情况。一个更强大的解决方案是：

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

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

这篇关于舍入整数除法（而不是截断）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！