bash - Bash 中的表达式递归级别超出错误

Question

我是 bash 脚本的新手，正在尝试做一些练习。当我试图用“完成”字符串停止程序时出现这样的错误。:

line 9: ((: finish: expression recursion level exceeded (error token is "finish").

问题是什么？我也想了解我的其他缺点。我的程序是:

#!/bin/bash
number=0
finish="finish"
temp=0
echo "Enter a number."
while true;
do
    read -r number
    if (( $number > $temp ))
    then
        temp=$number
    fi
    if [[ $number == $finish ]]
    then 
       break
    fi
done    
echo "Largest : $temp"  

Answer 1

引自@barmar's great answer到相关问题，该问题与@GordonDavisson 在他的评论中提到的相同问题:

When you use a variable in an arithmetic expression, but the value is not a number, the shell treats it as another expression to evaluate. So if the value is a variable name, it will get the value of that variable and use it. But in this case, you have it pointing to itself. So to evaluate a it has to evaluate $finish, and this keeps repeating infinitely.

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最简单的解决方案是为您的变量使用不同的名称 - 比如说 finish_string="finish" 而不是 finish="finish"。

此外，您可以使用正则表达式匹配来查看该值是否为数字(注意:不需要美元符号来扩展算术表达式 ((...)) 中的变量)，然后进行数字比较，然后进行正常的字符串比较，以查看值是否为“完成”:

if [[ $number =~ ^[0-9]+$ ]] && ((number > temp)); then
  temp=$number
elif [[ $number == $finish ]]; then
  break
fi

或者，您可以在进行数字比较之前明确检查用户是否输入了数字:

if [[ $number == $finish ]]; then
  break
else
  [[ $number =~ ^[0-9]+$ ]] || { echo "Enter a valid number or 'finish' to stop"; continue; }
  if ((number > temp)); then
    temp=$number
  fi
fi

使用 bash -x yourscript.sh 运行您的脚本以对其进行调试。

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相关:

• Expression recursion level exceeded

另见:

• How to use double or single brackets, parentheses, curly braces