Bash:将函数作为参数传递 [英] Bash: pass a function as parameter

问题描述

我需要在 Bash 中传递一个函数作为参数.例如下面的代码:

I need to pass a function as a parameter in Bash. For example, the following code:

function x() {
  echo "Hello world"
}

function around() {
  echo "before"
  eval $1
  echo "after"
}

around x

应该输出:

before
Hello world
after

我知道 eval 在那种情况下是不正确的，但这只是一个例子:)

I know eval is not correct in that context but that's just an example :)

有什么想法吗?

推荐答案

如果你不需要任何花哨的东西，比如延迟函数名或其参数的计算，你就不需要 eval:

If you don't need anything fancy like delaying the evaluation of the function name or its arguments, you don't need eval:

function x()      { echo "Hello world";          }
function around() { echo before; $1; echo after; }

around x

做你想做的.您甚至可以通过这种方式传递函数及其参数:

does what you want. You can even pass the function and its arguments this way:

function x()      { echo "x(): Passed $1 and $2";  }
function around() { echo before; "$@"; echo after; }

around x 1st 2nd

印刷品

before
x(): Passed 1st and 2nd
after

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