在 Bash 中舍入除数 [英] Round a divided number in Bash

问题描述

我将如何从两个相除的数字中取整结果，例如

How would I round the result from two divided numbers, e.g.

3/2

和我一样

testOne=$((3/2))

$testOne 包含1"，而它应该四舍五入为2"作为 3/2=1.5 的答案

$testOne contains "1" when it should have rounded up to "2" as the answer from 3/2=1.5

推荐答案

要在截断算术中进行四舍五入，只需将 (denom-1) 添加到分子即可.

To do rounding up in truncating arithmetic, simply add (denom-1) to the numerator.

示例，四舍五入:

N/2
M/5
K/16

示例，四舍五入:

(N+1)/2
(M+4)/5
(K+15)/16

要进行舍入到最近，请将 (denom/2) 添加到分子(一半将向上取整):

To do round-to-nearest, add (denom/2) to the numerator (halves will round up):

(N+1)/2
(M+2)/5
(K+8)/16

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