Can PHP dissect its own syntax?

Question

Can PHP dissect its own syntax? For example, I'd like to write a function that takes in an input like $object->attribute and says to itself:

OK, he's giving me $foo->bar, which means he must think that $foo is an object that has a property called bar. Before I try accessing bar and potentially get a 'Trying to get property of non-object' error, let me check whether $foo is even an object.

The end goal is to echo a value if it is set, and fail silently if not.

I want to avoid repetition like this:

<input value="<? if(is_object($foo) && is_set($foo->bar)){ echo $foo->bar; }?> "/>

...and to avoid writing a function that does the above, but has to have the object and attribute passed in separately, like this:

<input value="<? echoAttribute($foo,'bar') ?>" />

...but to instead write something which:

• preserves the object->attribute syntax
• is flexible: can also handle array keys or regular variables

Like this:

<input value="<? echoIfSet($foo->bar); ?> />
<input value="<? echoIfSet($baz['buzz']); ?> />
<input value="<? echoIfSet($moo); ?> />

But this all depends on PHP being able to tell me "what kind of thing am I asking for when I say $object->attribute or $array[$key]", so that my function can handle each according to its own type.

Is this possible?

Update

I got some good answers here and did some experimenting. Wanted to summarize.

• The answer to my original question appears to be "no," but we did find a way to accomplish was I was trying to do. Techpriester pointed out that I could pass the string '$foo->bar' to a function and have it parse that and eval() it. Not the approach I want to take, but deserves a mention.
• Peter Bailey pointed out that something like $foo->bar gets evaluated before getting passed to a function. I should have thought of this, but for some reason didn't. Thanks, Peter!
• Will Vousden showed that passing $foo->$bar by reference allows the function to evaluate it, rather than having it evaluated beforehand. His function shows that isset($foo->bar) will not complain if $foo is not an object, which I didn't know.

More interestingly, his example seems to imply that PHP waits to evaluate a variable until it absolutely has to.

If you're passing something by value, then of course PHP has to determine the value right then. Like this:

$foo->bar = 'hi';
somefunction($foo->bar); // same as somefunction('hi');

But if you're passing by reference, it can wait to determine the value when it actually needs to.

Like this (following the order in which things happen):

echo ifSet($foo->bar); // PHP has not yet evaluated $foo->bar...
function ifSet(&$somevar){ // ...because we're passing by reference (&$somevar)
  // Now we're inside the function, but it still hasn't evaluated $foo->bar; it 
  // just made its local variable, $somevar, point to the same thing as $foo->bar
  if(isset($somevar)){ // Right HERE is where it's evaluated - when we need it
    return $bar;
  }
}

Thanks to everyone who responded! If I've misstated something, please let me know.

Answer 1

You can pass it by reference:

<?php

function foo(&$bar)
{
    if (isset($bar))
    {
        echo "$bar\r\n";
    }
    else
    {
        echo "It's not set!\r\n";
    }
}

$baz = new stdClass;
$baz->test = 'test';
foo($baz->test);
foo($baz->test2);

$baz = array();
foo($baz['test3']);

?>

Note that this won't work if you try to access an object as an array or vice versa.

You shouldn't try to rely on something like this too much, though.