PHP a function in a functions argument?

Question

Is it possible to call a function as a function argument? I'm trying to get the value of the bar function and pass it to the foo function.

If I try to set the function to a variable

e.g.

$foobar = bar($x)

it will execute the function where ever the variable is found.

e.g.

function bar($x) {
    $x = 1;
    return 1;
}
function foo(x) {
    $x += $x;
    echo $x;
}

foo(bar($x));

Have you tried converting the pseudo-code to actual PHP code and running it? — BoltClock
– BoltClock, Commented Feb 12, 2011 at 20:53
What makes you think it shouldn't work? And how is that in any way related to functional programming? — user395760
– user395760, Commented Feb 12, 2011 at 21:02

Dan Grossman · Accepted Answer · 2011-02-12 20:54:57Z

4

Yes, you can do this in PHP, and virtually all high level programming languages. bar($x) will be evaluated, and its return value will be passed as the argument to foo().

answered Feb 12, 2011 at 20:54

Dan Grossman

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4 Comments

Dan Grossman Over a year ago

Thank you. I made it 10 years ago. PHP is pretty old. ;)

BoltClock Over a year ago

I made a similar avatar a while back, but the delimiters were in red, I used <?php and I didn't have a function call in the middle. Fun stuff!

Aaron Over a year ago

Dan what languages would you recommend?

Dan Grossman Over a year ago

Recommend for what? I recommend programming in whatever language you know best.

Spudley · Accepted Answer · 2011-02-12 20:55:32Z

0

x should be $x, other than that there's nothing wrong with your code. The answer is 'yes, you can use a function call as an argument in another function. It's pretty standard stuff.

answered Feb 12, 2011 at 20:55

Spudley

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