Just a quick and no doubt easy question. I'm pretty new to PHP and am looking through some existing code. I have tried to find the answer to my question on google, but to no avail.
Can somebody please let me know what the '&' sign before the parameter $var does??
function setdefault(&$var, $default="")
{
if (! isset($var))
{
$var = $default;
}
}
Passes it by reference.
Huh?
Passing by reference means that you pass the address of the variable instead of the value. Basically you're making a pointer to the variable.
http://us.php.net/language.references.pass
It means that the function gets the reference to the original value of the argument $var, instead of a copy of the value.
Example:
function add(&$num) { $num++; }
$number = 0;
add($number);
echo $number; // this outputs "1"
If add() would not have the ampersand-sign in the function signature, the echo would output "0", because the original value was never changed.
This means that you are passing a variable by reference http://ca3.php.net/language.references.pass. Simply this means the function is getting an the actual variable and not a copy of the variable. Any changes you make to that variable in the function will be mirrored in the caller.
It’s indicating that the parameter is passed by reference instead of by value.
& means pass-by-reference; what that code does is check whether the variable passed to the function actually exists in the global scope. Without the & it'd try to take a copy of the variable first, which causes an error if it doesn't exist.
Related
Is it considered wrong to pass a variable to a function where the variable passed has the same name as the variable in the function itself? I never see anyone do this, but wondered if there was a reason why (other than readability). Example:
$variable = "value";
function myFunction($variable) {
return $variable;
}
No there is not problem with doing this except some naming confusion. Because,
$variable = "value"; // this declaration belong to a global scope
// but $variable passing through as argument, makes it scope within that function declaration only
function myFunction($variable) {
return $variable;
}
You might have expect that if you call the function like myFunction() i.e. without any parameter, it will return you "value", but absolutely not, cause $variable passed as argument is not the global one.
But if you try to use the global $variable within the myFunction, then it will cause some conflict, that is why no one do this.
Aside from making variables a bit confusing to track, there is nothing wrong with this. They exist in different scopes.
It is up to you but vice versa, if you want to pass intentionally a variable into a function.. then you have to use this:
$variable = "value";
function myFunction($variable) {
global $variable;
return $variable;
}
echo myFunction('another value');
One more point. When creating functions you may set a default value for your params to avoid errors:
function myFunction($variable="default value") {
return $variable;
}
echo myFunction();
There is nothing wrong with it at all, and I would have thought it's quite common.
However, you should always be choosing the best name for a variable, or a parameter, in each situation, not applying one rule to your whole code.
For instance, in a database lookup function, it might be obvious what $id refers to, so it would be a sensible parameter name; but where you are calling it, you will likely have lots of IDs of various sorts, so a more descriptive variable name should be used. On the other hand, each time you build an SQL string, it's probably the only such variable in scope, so giving it a consistent name keeps your code tidy and easy to read.
A common observation is that all code is read more often than it is written (yes, even your personal website that you wouldn't dare share with the world). Variable names should be chosen with that in mind: what makes this code easy to read, understand, and maintain?
You can use global and add default value in param variable.
$variable = "value";
function myFunction($variable = '') {
global $variable;
return $variable;
}
func(CONST_A) should return 'CONST_A', func($name) should return $name
How to implement this func in PHP?
This doesn't seem possible easily.
You can use reflection to determine the parameters of the function, but that returns the variable names that the function expects, and if you're already inside the function, you kind of already know those.
debug_backtrace is the usual way of peeking at what called you, but it returns the values of the passed arguments, not the variable names or constants that the caller used when making the call.
However, what it does give you is the file name and line number of the caller, so you could open up the file and seek to that line and parse it out, but that would be very silly and you should not do this. I am not going to give you example code for this, as it is so utterly silly that you should not consider doing it ever.
The get_defined_vars thing is a hack and is not guaranteed to work either, and definitely won't work for constants, as get_defined_constants does that.
Try this:
<?php
function getVarConst($var)
{
if (isset($GLOBALS[$var]) // check if there is a variable by the name of $var
{
return $GLOBALS[$var]; // return the variable, as it exists
}
else if (defined($var)) // the variable didn't exist, check if there's a constant called $var
{
return constant($var); // return the constant, as it exists
}
else
{
return false; // return false, as neither a constant nor a variable by the name of $var exists
}
}
?>
This isn't possible.
You literally want the following right?
define('CONST_A', 'THIS COULD BE ANYTHING');
$name = 'who cares';
func(CONST_A); //returns 'CONST_A'
func($name); //returns '$name'
The function can't know that.
I suppose that reading the source code like Charles describes could get you this, but why?
Below is the code which outputs "15", why?
function zz(&$x){
$x = $x + 5;
}
$x = 10;
zz($x);
echo $x;
Please explain
Works as designed. By using & you pass $x by reference, meaning that anything the function does to the variable, will be done to the original $x that is set to 10.
If you used
function zz($x)
the original $x would stay at 10, because only the variable value is passed to the function.
you are passing the value as argument is not direct value of the variable but its passing By reference, so its giving you 15 as a output.
Thanks!
Because the function signature defines that the value passed to the function should be passed by reference.
If you don't know what that means, I suggest to read this paragraph on Wikipedia.
Adding a & means you are passing the $x variable by reference. The value outside is changed within the function, instead of a copy within the function being changed.
$x inside the function is a reference to the same value as $x outside your function.
When a function accepts a parameter with a "&", it's value is not copied into the new variable created inside the function's scope, but is a reference to the same value as the argument that was given.
See here.
Using & Ampersand: Passing by Reference mets the purpose in the function.
Its simply alter the original variable and return it again to the same variable name with its new value assigned.
Sorry for such a lame title, but I just had no idea what to put there, hope you understand it. Plus, I have no idea if similar question has been asked before, because I don't know the proper keywords for it - therefore couldn't google it too.
Basicly... When looking at preg_match_all();, they got this matches parameter that will create new array defined in function, and give us the ability to access it after function execution.
And the question is.. How can I implement such a feature in my own function? But that it could create single variable and/or array.
Thanks in advance!
preg_match_all() accepts a reference to an array, which in its own scope is called $matches. As seen in the function prototype:
array &$matches
If you call the function and pass in a variable, if it does not already exist in the calling scope it will be created. So in your user-defined function, you accept a parameter by reference using &, then work with it inside your function. Create your outer-scope variable by simply declaring it in your function call, like you the way you call preg_match_all() with $matches.
An example:
function foo(&$bar) {
$bar = 'baz';
}
// Declare a variable and pass it to foo()
foo($variable);
echo $variable; // baz
I think you are referring to function parameters passed by reference, are you not?
function putValInVar(&$myVar, $myVal){
$myVar = $myVal;
}
$myVar = 1;
putValInVar($myVar, 2);
echo $myVar; // outputs '2', but will output '1' if we remove the '&' //
By default function arguments in PHP are passed by value. This means that new variables are created at each function call and those variables will exist only inside the function, not affecting anything outside it.
To specify that an argument should be used by reference the syntax is to append an & before declaring it in the function header. This will instruct PHP to use the passed variable inside the function rather than creating a copy of it.
Exception: Objects are always passed by reference. (Well... Not really, but it's complicated. See the comment thread for more info.)
I think what you are asking for is passing-by-reference. What preg_match_all basically does to "create" an array variable outside its scope is:
function preg_match_all($foo, $bar, & $new_var) {
$new_var = array(1,2,3);
}
The crucial point here is & in the function definition. This allows you to overwrite variables in the outer scope when passed.
Stylistically this should be used with care. Try to return arrays or results instead of doing it via reference passing.
Like this:
$myvariable = runfunction();
function runfunction() {
//do some code assign result to variable (ie $result)
return $result;
}
Or
global $result;
function runfunction() {
global $result;
$result = 'something';
}
I'm pretty sure the answer to this question is no, but in case there's some PHP guru
is it possible to write a function in a way where invalid arguments or non existent variables can be passed in and php will not error without the use of '#'
Much like empty and isset do. You can pass in a variable you just made up and it won't error.
ex:
empty($someBogusVar); // no error
myHappyFunction($someBogusVar); // Php warning / notice
You don't get any error when a variable is passed by reference (PHP will create a new variable silently):
function myHappyFunction(&$var)
{
}
But I recommend against abusing this for hiding programming errors.
Summing up, the proper answer is no, you shouldn't (see caveat below).
There are workarounds already mentioned by many people in this thread, like using reference variables or isset() or empty() in conditions and suppressing notices in PHP configuration. That in addition to the obvious workaround, using #, which you don't want.
Summarizing an interesting comment discussion with Gerry: Passing the variable by reference is indeed valid if you check for the value of the variable inside the function and handle undefined or null cases properly. Just don't use reference passing as a way of shutting PHP up (this is where my original shouldn't points to).
You can do this using func_get_args like so:
error_reporting(E_ALL);
ini_set('display_errors', 1);
function defaultValue() {
$args = func_get_args();
foreach($args as $arg) {
if (!is_array($arg)) {
$arg = array($arg);
}
foreach($arg as $a) {
if(!empty($a)) {
return $a;
}
}
}
return false;
}
$var = 'bob';
echo defaultValue(compact('var'), 'alpha') . "\n"; //returns 'bob'
echo defaultValue(compact('var2'), 'alpha') . "\n"; //returns 'alpha'
echo defaultValue('alpha') . "\n"; //return
echo defaultValue() . "\n";
This func goes one step further and would give you the first non empty value of any number of args (you could always force it to only take up to two args but this look more useful to me like this).
EDIT: original version didn't use compact to try and make an array of args and STILL gave an error. Error reporting bumped up a notch and this new version with compact is a little less tidy, but still does the same thing and allows you to provide a default value for non existent vars.
There are valid cases where checking becomes cumbersome and unnessesary.
Therfore i've written this little magic function:
/**
* Shortcut for getting a value from a possibly unset variable.
* Normal:
* if (isset($_GET['foo']) && $_GET['foo'] == 'bar') {
* Short:
* if (value($_GET['foo']) == 'bar') {
*
* #param mixed $variable
* #return mixed Returns null if not set
*/
function value(&$variable) {
if (isset($variable)) {
return $variable;
}
}
It doesn't require any changes to myHappyFunction().
You'll have to change
myHappyFunction($someBogusVar);
to
myHappyFunction(value($someBogusVar));
Stating your intent explicitly. which makes it good practice in my book.
No, because this isn't really anything to do with the function; the error is coming from attempting to de-reference a non-existent array key. You can change the warning level of your PHP setup to surpress these errors, but you're better off just not doing this.
Having said that, you could do something like
function safeLookup($array, $key)
{
if (isset($array, $key))
return $array[$key];
return 0;
}
And use it in place of array key lookup
defaultValue(safeLookup($foo, "bar"), "baz);
Now I need to take a shower :)
is it possible to write a function in a way where invalid arguments or non existent variables can be passed in and php will not error without the use of '#'
Yes you can!
porneL is correct [edit:I don't have enough points to link to his answer or vote it up, but it's on this page]
He is also correct when he cautions "But I recommend against abusing this for hiding programming errors." however error suppression via the Error Control Operator (#) should also be avoided for this same reason.
I'm new to Stack Overflow, but I hope it's not common for an incorrect answer to be ranked the highest on a page while the correct answer receives no votes. :(
#Brian: I use a trinary operation to do the check for me:
return $value ? $value : $default;
this returns either $value OR $default. Depending upon the value of $value. If it is 0, false, empty or anything similar the value in $default will be returned.
I'm more going for the challenge to emulate functions like empty() and isset()
#Sean That was already answered by Brian
return isset($input) ? $input : $default;
Sean, you could do:
$result = ($func_result = doLargeIntenseFunction()) ? $func_result : 'no result';
EDIT:
I'm sure there could be a great
discussion on ternary operators vrs
function calls. But the point of this
question was to see if we can create a
function that won't throw an error if
a non existent value is passed in
without using the '#'
And I told you, check it with isset(). A ternary conditional's first part doesn't check null or not null, it checks true or false. If you try to check true or false on a null value in PHP, you get these warnings. isset() checks whether a variable or expression returns a null value or not, and it returns a boolean, which can be evaluated by the first part of your ternary without any errors.
I'm sure there could be a great discussion on ternary operators vrs function calls. But the point of this question was to see if we can create a function that won't throw an error if a non existent value is passed in without using the '#'
While the answer to the original question is "no", there is an options no one has mentioned.
When you use the # sign, all PHP is doing is overriding the error_reporting level and temporarily setting it to zero. You can use "ini_restore('error_reporting');" to set it back to whatever it was before the # was used.
This was useful to me in the situation where I wanted to write a convenience function to check and see if a variable was set, and had some other properties as well, otherwise, return a default value. But, sending an unset variable through caused a PHP notice, so I used the # to suppress that, but then set error_reporting back to the original value inside the function.
Something like:
$var = #foo($bar);
function foo($test_var)
{
ini_restore('error_reporting');
if(is_set($test_var) && strlen($test_var))
{
return $test_var;
}
else
{
return -1;
}
}
So, in the case above, if $bar is not set, I won't get an error when I call foo() with a non-existent variable. However, I will get an error from within the function where I mistakenly typed is_set instead of isset.
This could be a useful option covering what the original question was asking in spirit, if not in actual fact.
If you simply add a default value to the parameter, you can skip it when calling the function. For example:
function empty($paramName = ""){
if(isset($paramName){
//Code here
}
else if(empty($paramName)){
//Code here
}
}
With a single line, you can acomplish it: myHappyFunction($someBogusVar="");
I hope this is what you are looking for. If you read the php documentation, under default argument values, you can see that assigning a default value to an function's argument helps you prevent an error message when using functions.
In this example you can see the difference of using a default argument and it's advantages:
PHP code:
<?php
function test1($argument)
{
echo $argument;
echo "\n";
}
function test2($argument="")
{
echo $argument;
echo "\n";
}
test1();
test1("Hello");
test1($argument);
$argument = "Hello world";
test1($argument);
test2();
test2("Hello");
test2($argument);
$argument = "Hello world";
test2($argument);
?>
Output for test1() lines:
Warning: Missing argument 1 for test1() .
Hello.
.
Hello world.
Output for test2() lines:
.
Hello.
Hello world.
This can also be used in combination to isset() and other functions to accomplish what you want.
And going further up the abstraction tree, what are you using this for?
You could either initialize those values in each class as appropriate or create a specific class containing all the default values and attributes, like:
class Configuration {
private var $configValues = array( 'cool' => 'Defaultcoolval' ,
'uncool' => 'Defuncoolval' );
public setCool($val) {
$this->configValues['cool'] = $val;
}
public getCool() {
return $this->configValues['cool'];
}
}
The idea being that, when using defaultValue function everywhere up and down in your code, it will become a maintenance nightmare whenever you have to change a value, looking for all the places where you've put a defaultValue call. And it'll also probably lead you to repeat yourself, violating DRY.
Whereas this is a single place to store all those default values. You might be tempted to avoid creating those setters and getters, but they also help in maintenance, in case it becomse pertinent to do some modification of outputs or validation of inputs.