如何在函数内访问调用方的命令行参数? [英] How to access command line arguments of the caller inside a function?

问题描述

我正在尝试用bash编写一个函数，该函数将访问脚本的命令行参数，但是将它们替换为该函数的位置参数.如果未显式传递命令行参数，函数有什么方法可以访问它们?

I'm attempting to write a function in bash that will access the scripts command line arguments, but they are replaced with the positional arguments to the function. Is there any way for the function to access the command line arguments if they aren't passed in explicitly?

# Demo function
function stuff {
  echo $0 $*
}

# Echo's the name of the script, but no command line arguments
stuff

# Echo's everything I want, but trying to avoid
stuff $*

推荐答案

我对bash参考手册的阅读表明，这些东西是在BASH_ARGV中捕获的， 尽管它经常谈论堆栈".

My reading of the bash ref manual says this stuff is captured in BASH_ARGV, although it talks about "the stack" a lot.

#!/bin/bash

function argv {
    for a in ${BASH_ARGV[*]} ; do
      echo -n "$a "
    done
    echo
}

function f {
    echo f $1 $2 $3
    echo -n f ; argv
}

function g {
    echo g $1 $2 $3
    echo -n g; argv
    f
}

f boo bar baz
g goo gar gaz

保存在f.sh中

$ ./f.sh arg0 arg1 arg2
f boo bar baz
farg2 arg1 arg0 
g goo gar gaz
garg2 arg1 arg0 
f
farg2 arg1 arg0 

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