bash：变量内的算术表达式

Question

我有一个简单的代码来基于简单的算术等式在循环中分配变量

 # assign initial value
    restr_start='25'
    # assign a new variable, which is a number that will decrease initial value by 5
    # keeping always the value of previous variable as restr_prev 
    for step in {1..4}; do
      let "restr=(${restr_start} - (5 * ${step}))"
      let "restr_prev=(${restr} + (5 * ${step}))"
      echo this is $restr current restart
      echo this is $restr_prev previous restart
    done

从这个脚本中我期望得到：

this is 20 current restart
this is 25 previous restart
this is 15 current restart
this is 20 previous restart
this is 10 current restart
this is 15 previous restart
this is 5 current restart
this is 10 previous restart

然而我实际上有什么

this is 20 current restart
this is 25 previous restart
this is 15 current restart
this is 25 previous restart
this is 10 current restart
this is 25 previous restart
this is 5 current restart
this is 25 previous restart

为什么 $restr_prev 通常是不变的？我如何修改代码，例如使用某些东西代替 let？

Answer 1

这是一个数学问题，而不是 bash 代码的问题。看$restr_prev的公式：

restr_prev= ${restr} + (5 * ${step})

对于步骤1，公式计算为20 + 5 * 1 = 25，对于步骤2，公式导致15 + 5 * 2 = 25，依此类推...

为了获得您实际期望的结果，您只需将5 添加到restr 值。因此，脚本中的相应行应如下所示：

let "restr_prev=(${restr} + 5)"

正如 cmets 中已经建议的那样，您应该使用 $((expression)) 而不是 let 进行算术扩展，因为后者是内置的 bash 并且不包含在 POSIX standard 中。听取建议导致以下代码：

#!/bin/bash

# assign initial value
restr_start='25'
# assign a new variable, which is a number that will decrease initial value by 5
# keeping always the value of previous variable as restr_prev 
for step in {1..4}; do
    restr=$((restr_start - (5 * step)))
    restr_prev=$((restr + 5))
    echo "this is $restr current restart"
    echo "this is $restr_prev previous restart"
done