此问题已在此处有答案：

Propagate all arguments in a Bash shell script（12个回答）
Accessing bash command line args $@ vs $*（5个答案）
昨天关门了。
我正在编写一个简单的bash脚本来为git status添加一些额外的功能。目前，我正在通过将以下脚本（名为git）添加到我的路径来实现这一点：

#!/bin/bash

# Simple script to automatically configure user info between multiple repositories
#  by overloading the functionality of 'git status'
if [[ "$1" == "status" ]]; then
    # do something extra, then run vanilla 'git status'
    /usr/bin/git status
else
    # behave normally
    /usr/bin/git $@
fi

当我输入git <any args>时，Bash会首先找到我的脚本git，所以它会运行它。我遇到的问题是else子句。显然，在这样的脚本中运行/usr/bin/git $@与从命令行运行它是不一样的。例如，当我运行git commit时：

$ git commit -m "Init HW4 Rust project" # runs the else branch of my script
error: pathspec 'HW4' did not match any file(s) known to git
error: pathspec 'Rust' did not match any file(s) known to git
error: pathspec 'project' did not match any file(s) known to git
$ /usr/bin/git commit -m "Init HW4 Rust project" # runs my standard git installation
On branch main
Your branch is ahead of 'origin/main' by 2 commits.
  (use "git push" to publish your local commits)

nothing to commit, working tree clean

我怀疑这与$@变量有关。据我所知，$@表示Bash脚本接收的所有命令行参数，除了第0个参数（命令）。这与显式键入参数有什么不同？