Questions about function return values

I still don’t know much about function return values. In the following example, if jj() is called, a 1 will be printed directly. Why is a null printed? It seems that the function return value is not used here. If you call this function, don’t you just execute $ Is it enough to just print it out? Why is it still related to return value?

<code>    function jj(){
        $a=1;
        echo $a;
    }
    $b=jj();
    var_dump($b);</code>

Reply content:

I still don’t know much about function return values. In the following example, if jj() is called, a 1 will be printed directly. Why is a null printed? It seems that the function return value is not used here. If you call this function, don’t you just execute $ Is it enough to just print it out? Why is it still related to return value?

<code>    function jj(){
        $a=1;
        echo $a;
    }
    $b=jj();
    var_dump($b);</code>

Print is only displayed on the screen. If you use this value, you must use return to receive it elsewhere. So here your function does not return anything, it just prints 1, so the variable b does not get any value, so it is empty.

Because your function does not have a return statement, null is returned after execution.
Your function execution process:

1. $b = jj(); Call jj() to assign value to $a, print $a, no return (return is null)

2. var_dump $b, $b is assigned the return value of jj(), so it is null

3. After the entire execution is completed, the printed 1 (value of $a), null (value of $b) are left

<code>function jj(){
        $a=1;
        echo $a;
        return $a;
    }
    $b=jj();
    var_dump($b);</code>

This is probably the output you expect

Your jj function directly outputs $a without return, so $b is empty

<code>$b=jj();//输出1 直接调用函数jj
var_dump($b);//输出null  打印函数？并没用return值

加了return之后的结果如下：
1
D:\WWW\demo\demo\demo.php:15:int 1
</code>

echo is direct output, return returns the location where it is called