在Bash或Shell脚本中转发函数声明？

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问题描述

在 bash 中是否存在这样的事情，或者至少有类似于前向声明的类似（解决方法），例如C / C ++中众所周知的？

Is there such a thing in bash or at least something similar (work-around) like forward declarations, well known in C / C++, for instance?

或者有这样的事情，因为例如它总是以一次执行（一行一行）？

Or there is so such thing because for example it is always executed in one pass (line after line)?

如果没有前向声明，我应该怎么做才能让我的脚本更易于阅读。这是相当长的，开头的这些函数定义，与全局变量混合在一起，使得我的脚本看起来丑陋难以阅读/理解）？我要求了解一些此类案例的知名/最佳实践。

If there are no forward declarations, what should I do to make my script easier to read. It is rather long and these function definitions at the beginning, mixed with global variables, make my script look ugly and hard to read / understand)? I am asking to learn some well-known / best practices for such cases.

例如：

# something like forward declaration
function func

# execution of the function
func

# definition of func
function func
{
    echo 123
}

推荐答案

好的问题。对于我的大部分脚本，我都使用这种模式：

Great question. I use a pattern like this for most of my scripts:

#!/bin/bash

main() {
    foo
    bar
    baz
}

foo() {
}

bar() {
}

baz() {
}

main "$@"

您可以从上至下阅读代码，但直到最后一行才会开始执行。通过将$ @传递给main（），您可以访问命令行参数 $ 1 ， $ 2 ，等等。就像你平常那样。

You can read the code from top to bottom, but it doesn't actually start executing until the last line. By passing "$@" to main() you can access the command-line arguments $1, $2, et al just as you normally would.

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declarations