为什么变量分配中的空格会导致Bash错误? [英] Why does a space in a variable assignment give an error in Bash?

问题描述

#!/bin/bash

declare -r NUM1=5
NUM2 =4                # Line 4

num3=$((NUM1 + NUM2))
num4=$((NUM1 - NUM2))
num5=$((NUM1 * NUM2))
num6=$((NUM1 / NUM2))  # Line 9

echo "$num3"
echo $((5**2))
echo $((5%4)) 

我正在使用此bash脚本，当我运行该脚本时，出现了错误

I am using this bash script, and when I was running the script, I got the error

./bash_help 
./bash_help: line 4: NUM2: command not found
./bash_help: line 9: NUM1 / NUM2: division by 0 (error token is "NUM2")
5
25
1

所以我将代码更改为此，错误消失了.

So I have changed the code to this and the error was gone.

#!/bin/bash

declare -r NUM1=5
NUM2=4

num3=$((NUM1 + NUM2))
num4=$((NUM1 - NUM2))
num5=$((NUM1 * NUM2))
num6=$((NUM1 / NUM2))

echo "$num3"
echo $((5**2))
echo $((5%4)) 

为什么在给变量赋值时不能使用空格?习惯上使用空格以提高代码的可读性.谁能解释一下?

Why can't we use spaces when we assign a value to a variable? It is convention to use spaces for better readability of the code. Can anyone explain this?

推荐答案

这不是bash中的约定(或更常见的是POSIX系列shell).

It's not a convention in bash (or, more generally, POSIX-family shells).

至于为什么"，这是因为各种错误的处理方式都具有作为命令的有效含义.如果对NUM2 = 4进行了赋值，则无法在不加引号的情况下将=作为文字参数传递. 因此，任何此类更改都将是向后不兼容的，而不是将其放置在未定义的空间中(需要保留POSIX sh标准的扩展，以避免构成对该标准的违反).

As for "why", that's because the various ways of doing it wrong all have valid meanings as commands. If you made NUM2 = 4 an assignment, then you couldn't pass = as a literal argument without quoting it. Consequently, any such change would be backwards-incompatible, rather than being placed in undefined space (where extensions to the POSIX sh standard need to live to avoid constituting violations of that standard).

NUM2= 4 # runs "4" as a command, with the environment variable NUM2 set to an empty string
NUM2 =4 # runs "NUM2" as a command, with "=4" as its argument
NUM2 = 4 # runs "NUM2" as a command, with "=" as its first argument, and "4" as another

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