I want to increase $x by 1:
$x = 1;
function addOne($x) {
$x++;
return $x;
}
$x = addOne($x);
Is there a way to do this with references, so I don't need to return $x, I can just write addOne($x)?
2 Answers
This is what you're looking for, a by-ref parameter indicated by &.
$x = 1;
function addOne(&$x) {
$x++;
}
addOne($x);
Some notes:
- By-ref parameters require that the value passed in not be a literal. Given my example above,
addOne(5)would throw a fatal exception - References are not needed for objects (including stdClass objects), as all objects are passed by reference as of PHP 5.
- References ARE needed for arrays, as arrays in PHP are not treated as objects
- If you want a return value of a function passed by reference, you would indicate the reference on the function name (e.g.
function &foo()).
More info on references: http://php.net/manual/en/language.references.php
4 Comments
kuujo
Was gonna mention giving some info on references in your answer but you beat me to it.
Don P
Is there a difference between using the reference in the function declaration, versus in the function call? Like
function addOne($x) but then just call it with addOne(&$x)?
FloydThreepwood
@DonnyP This is not C. Calling it that way will issue a warning.
bwoebi
@FloydThreepwood depends on the version of PHP. In newer versions than PHP 5.3, calltime-pass-by-ref will cause a fatal error.
$x = 1;
function addOne(&$x) {
$x++;
}
addOne($x);
The & sign shows that it takes the parameter by reference. So it increments $x in the function and it will also affect the $x variable in the calling scope.
See also: http://php.net/references for a quick overview about them.
Comments
$x++;:D