Bash $ @变量失败

Question

我已经好几次这样做了,它似乎永远不会正常工作.有谁能解释为什么？

function Foobar
{
  cmd -opt1 -opt2 $@
}

这是什么应该做的是让这个电话Foobar做同样的事情与调用cmd,但有一些额外的参数(-opt1和-opt2,在这个例子中).

不幸的是,这不能正常工作.如果所有参数都缺少空格,它就可以了.但是如果你想要一个带空格的参数,你可以用引号写出来,而Bash有助于剥离引号,打破命令.如何防止这种错误行为？

Answer 1

$@在替换参数值之后,你需要加倍引用以防止bash执行不需要的解析步骤(单词拆分等):

function Foobar
{
  cmd -opt1 -opt2 "$@"
}

从bash手册页的特殊参数部分编辑:

@   Expands to the positional parameters, starting from  one.   When
    the  expansion  occurs  within  double  quotes,  each  parameter
    expands to a separate word.  That is, "$@" is equivalent to "$1"
    "$2"  ...   If the double-quoted expansion occurs within a word,
    the expansion of the first parameter is joined with  the  begin-
    ning  part  of  the original word, and the expansion of the last
    parameter is joined with the last part  of  the  original  word.
    When  there  are no positional parameters, "$@" and $@ expand to
    nothing (i.e., they are removed).