如何在带参数的函数中调用变量函数？(How can I call a variable function within a function with parameters?)

 我试图在函数内调用一个变量函数，但它没有按预期工作。 
 
 这是我的代码： 
 
$arg1 = "Beef";
$arg2 = "Potatoes";

$string1 = "Tomatoes";
$string2 = "Onions";

function func1 ($arg1, $arg2){
    echo $arg1.' '.$arg2;
}

$var1 = "func1";

function func2($string1, $string2, $var1) {
    echo $string1." & ".$string2;
    $var1($arg1, $arg2);
}

func2($string1, $string2, $var1);
 
 当我调用func2 ，它只回应“Tomatoes＆Onions”Isnt $arg1和$arg2 global？ 为什么在函数func2没有看到它们？ 如何在func2调用$var1及其参数？ 

I am trying to call a variable function within a function but it is not working as expected.
 
Here is my code: 
 
$arg1 = "Beef";
$arg2 = "Potatoes";

$string1 = "Tomatoes";
$string2 = "Onions";

function func1 ($arg1, $arg2){
    echo $arg1.' '.$arg2;
}

$var1 = "func1";

function func2($string1, $string2, $var1) {
    echo $string1." & ".$string2;
    $var1($arg1, $arg2);
}

func2($string1, $string2, $var1);
 
When I call func2, it only echoes "Tomatoes & Onions" Isnt $arg1 and $arg2 global? Why are they not seen inside the function func2? How can I call $var1 inside func2 along with its arguments?

原文：