<!doctype html>
<html>
<head>
</head>
<body>
<?php
function distance($x,$y){
distance(10,10);
$d = $x * $y;
return $d;
echo $d; //not working
}
echo $d;
if ($d > 5){ //notworking
echo "grats";
}
?>
</body>
</html>
my code on line 14 and 11 is not working, line 11 wont output $d. line 14 wont call upon returned value $d.
It wont work, becase you haven't called the function. Use this instead:
<?php
function distance($x,$y)
{
return ($x * $y);
}
$d = distance(10, 10);
if ($d > 5) {
echo "grats";
}
?>
The return statement leaves the function. Use
echo distance(); to display the function return value or
$d = distance(); for assigning the value to a var
This may help in understanding return: http://www.php.net/manual/en/function.return.php
You have multiple problems.
return $d;
echo $d; < not working
The above does not work because when you call return in a function the function immediately stops executing. Any code after the return statement will not execute.
distance(10,10);
The above line does nothing in our function except call your function recursively forever. I don't know why your code does not throw an error for this. You can safely remove this as it does nothing constructive.
if ($d > 5){ <not working
The above code does not work because you never call your function to populate $d with a value. That line needs to look like this:
$d = distance(10, 10);
if ($d > 5){