为什么 `command -v` 退出代码在 bash 函数内部和外部的行为不同？

Question

考虑以下 bash 脚本：

#!/bin/bash

# List of required commands
required_commands=('git', 'curl', 'sdfsdf', 'jq')

# this will correctly exit if the 'git' command is missing
cc1='git'
command -v '$cc1' >/dev/null 2>&1 || { echo "Missing '$cc1' cannot run."; exit 1; } 

# this function, when called below will have an error exit code for each
missing_commands=()
check_command() {
    command -v "$1" >/dev/null 2>&1 || missing_commands+=("$1")
}

# check each required command
for cmd in "${required_commands[@]}"; do
    check_command "$cmd"
done

# Check if any commands are missing
if [ ${#missing_commands[@]} -eq 0 ]; then
    echo "All required commands are available. Starting the script..."
    # Your script continues here
else
    echo "The following commands are required but not installed:"
    for missing_cmd in "${missing_commands[@]}"; do
        echo "  - $missing_cmd"
    done
    exit 1
fi

# continue if all commands are available
echo "All required commands are available. Starting."

不幸的是，当在

check_command()

函数内部调用时，

command -v "$1" >/dev/null 2>&1

的退出代码始终非零？这看起来很不寻常，我不明白有什么区别。

这里的目标操作系统是 MacOS Sonoma 14.2.1 (23C71)，在普通终端中的 ZSH 中运行。

Answer 1

感谢 @Cyrus 和 https://shellcheck.net 的帮助，答案原来是我自己的愚蠢......逗号不用作 bash 数组中的分隔符，所以下面的行：

required_commands=('git', 'curl', 'sdfsdf', 'jq')

结果有四个项目，每个项目都以

,

字符结尾。哎呀。