将void方法的返回值赋给变量有什么意义？ [关闭]

Question

class test {
    public function hello() {
        echo "hello!";
    }
    public function assign() {
        $var = $this->hello();
        $this->hello(); //both seem to be doing the same thing
    }
}

This is a simplified example of a piece of code that I am talking about, which is here on line 43: https://eval.in/988393

What is the point of a assigning a method call (I guess this is how it is called) to a variable and what is the explanation of this causing it to get run (it does not just assign it but it executes the code).

PS: This is actually related to Iterators like here: http://php.net/manual/en/class.iterator.php . Somebody told me this:

"The point with iterators is that you do not care how the methods are called, you only care about each single method separatly".

What did this person mean?

Answer 1

Basically, in your example and the linked example of the iterator, assigning the return value is unnecessary.

• In your example it is because there is no return value. In other words, the function returns void which will cause $var to be null. In general, if a function contains no return statement at all or only a return; (note that there is no variable being returned), it is useless to store the result in a variable.
• In the iterator example it is unnecessary, because the value isn't used. But the $current variable does still contain the actual value of the current iteration - it is not null (except when the array contains null, obviously). Actually calling $this->current() wouldn't be necessary at all in the iterator implementation - it looks like it is done solely for the purpose of getting the nice output of echo "current: $var "; when checking for the validity.

There have been other questions asked about void return types in the past as well, e.g. Void as return type.

---

By the way, if your question was also about storing a function in a variable (and not its result), then you probably meant something like this:

function foo() {
    echo 'bar';
}

$var = 'foo';
$var(); // outputs: bar

But be careful with these kinds of function invokations, there are quite some pitfalls coming with them.