bash - IFS 如何在 Bash 中工作？

Question

#!/bin/bash
# This question is from advanced bash scripting guide section 5.1

echo
var="'(]\\{}\$\""
IFS='\'
echo $var

    # output is '(] {}$"     
    # \ converted to space. Why?

echo "$var"      

    # output is '(]\{}$"      
    # special meaning of \ used, \ escapes \ $ and " RIGHT?

echo
var2="\\\\\""
echo $var2       

    # output is   "         
    # \ converted to space. Why?

echo

    # But ... var2="\\\\"" is illegal. Why?

var3='\\\\'
echo "$var3"     # \\\\

    # Strong quoting works, though. Why?

Answer 1

IFS='\'
echo $var        

    # o/p is '(] {}$"     

    # \ converted to space. Why?

因为您告诉 shell 反斜杠是字段分隔符，并且由于您在回显时没有引用$var，所以它会根据 IFS进行分词。

echo "$var"      

    # o/p is '(]\{}$"      
    # special meaning of \ used, \ escapes \ $ and " RIGHT ?

您在这里引用$var，因此不会对其执行分词。您的输出正是您告诉 shell 的var结果。IE'(]\{}$"

var2="\\\\\""

echo $var2       

    # o/p is   "        
    # \ converted to space. Why?

看第一个答案

# But ... var2="\\\\"" is illegal. Why?

因为每对反斜杠都构成一个文字反斜杠，并且没有剩余的反斜杠可以逃脱第二个双引号。shell 不知道如何处理 3 个双引号。

echo "$var3"     # \\\\

    # Strong quoting works, though. Why ?

请参阅有关分词的第二个答案

请注意，您还可以使用字符串文字语法$''vis var=$'\'(]\{}$"'，它只需要您转义单引号