CODEWITHSUNDEEP
php
Notty
Notty

Reputation:

Why does "return fopen" fail?

I have a function:

function open($file){
    return fopen($file, 'w');
}

This is then called by:

function write($file,$text){
    $h = $this->open($file);
    fwrite($h,$text);
}

This doesn't work. It returns that fwrite was given an invalid resource stream.

This:

function open($file){
    $h = fopen($file, 'w');
    return $h;
}

Works fine, but I can't figure out why assigning a variable first works and directly returning fopen() doesn't.

Upvotes: 1

Views: 348

Answers (3)

Apaella
Apaella

Reputation:

declare a

var $file

then

$this->file = fopen(...)

return $this->file;

this will works because the $file variable has still a reference.

Upvotes: 0

strubester
strubester

Reputation: 285

It's probably just because you are working in the scope of an object, so it cleans up the resource stream too early - since it passes a resource stream byref, if you have a variable set, its byref'ing the variable instead of trying to do it to the resource stream - so it'll work.

Upvotes: 0

Tim
Tim

Reputation: 60150

Does it have something to do with the fact that you're within an object? The following script works for me:

<?php
        function open($file) {
                return fopen($file, 'w');
        }

        function write($file, $text) {
                $h = open($file);
                fwrite($h, $text);
        }

        write("test.txt", "hello\n");

?>

I'm running PHP 5.2.8 on Mac OS X 10.5.7.

Upvotes: 1

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