将带有空格的字符串作为参数传递给Bash函数

Question

给出以下字符串变量

VAR="foo         bar"

我需要将它传递给bash函数，并像往常一样通过$1访问它。到目前为止，我还没弄清楚如何做到这一点：

#!/bin/bash
function testfn(){
    echo "in function: $1"
}
VAR="foo         bar"
echo "desired output is:"
echo "$(testfn 'foo           bar')"
echo "Now, what about a version with \$VAR?"
echo "Note, by the way, that the following doesn't do the right thing:"
echo $(testfn "foo           bar") #prints: "in function: foo bar"

Answer 1

Bash很聪明，双引号对匹配$( ... )结构的内部或外部。

因此，echo "$(testfn "foo bar")"有效，testfn函数的结果只会被视为echo内部命令的单个参数。