Question

I am learning php and read this example in this http://www.php.net/manual/en/language.types.boolean.php tutorial,
I want to understand what occur when assigning the return value of method to a variable, why it may change?? please see my questions in the code.

<?php

public function myMethod()
{
return 'test';
}

public function myOtherMethod()
 {
 return null;
 }

if($val = $this->myMethod())
{
  // $val might be 1 instead of the expected 'test'  
                                                    ** why it may returns 1??**

}

if( ($val = $this->myMethod()) )
{
// now $val should be 'test'
}

// or to check for false
if( !($val = $this->myMethod()) )                    **what happens here????**
{
 // this will not run since $val = 'test' and equates to true
 }

// this is an easy way to assign default value only if a value is not returned:

if( !($val = $this->myOtherMethod()) )             **what happens here????**
{
$val = 'default'
}

?>

Answer 1

In this case:

if($val = $this->myMethod())
{
  // $val might be 1 instead of the expected 'test'  

}

I don’t think that’s true. $val should be ‘test’ here. Maybe in older versions of PHP there could have been a bug.

if(!($val = $this->myMethod()))
{
 // this will not run since $val = 'test' and equates to true
}

Here myMethhod() is executed and returns ‘test’ which is assigned to $val. Then the result of that assignment is boolean negated. Since the string ‘test’ evaluates to true, !(‘test’) evalutes to false and the if statement doesn’t run.

if(!($val = $this->myOtherMethod()))
{
$val = 'default';
}

This is the opposite case. $val becomes null. And null evaluates to boolean false, so !(null) evaluates to true and the code in the block executes. So after this code runs $val contains ‘default’; This poster is showing this as a way of assigning a default value to $val in the case that $this->myOtherMethod() fails to return anything useful.