php: Declare arguments type of a Function

Question

I'm trying to make a function with declared argument types, to quickly check if they are in the right format, but when it returns a string I this error:

Catchable fatal error: Argument 2 passed to myfunction() must be an instance of string, string given, called in path_to_file on line 69 and defined in path_to_file on line 49

Example

function myfunction( array $ARRAY, string $STRING, int $INTEGER ) { 
    return "Args format correct"; 
}
myfunction(array("1",'2','3','4'), "test" , 1234);

Where is the mistake?

Solution 1

According to the PHP5 documentation:

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported.

Since string and int are not classes, you can't "type-hint" them in your function.

As of PHP 7.0 declaring argument type as string, int, float, bool is supported.

Solution 2

This maybe useful for anyone who see this post since the availability of PHP 7

With PHP 7, its now possible to declare types. You can refer the following link for more information.

http://php.net/manual/en/functions.arguments.php#functions.arguments.type-declaration

function(string $name, bool $is_admin) {
    //do something
}

Solution 3

According to PHP Manual, you can do that for array on PHP 5.1 and beyond and for string and int types on PHP 7 and beyond. Take a look:

• Class/interface name The parameter must be an instanceof the given class or interface name. PHP 5.0.0
• self The parameter must be an instanceof the same class as the one the method is defined on. This can only be used on class and instance methods. PHP 5.0.0

• array The parameter must be an array. PHP 5.1.0

• callable The parameter must be a valid callable. PHP 5.4.0
• bool The parameter must be a boolean value. PHP 7.0.0
• float The parameter must be a floating point number. PHP 7.0.0

• int The parameter must be an integer. PHP 7.0.0

• string The parameter must be a string. PHP 7.0.0

• iterable The parameter must be either an array or an instanceof Traversable. PHP 7.1.0

Solution 4

You can do something like this which has always worked for me

for string

function setData($Name=""){ }

this forces the name to be a string, it doesn't check if its a string

for numeric values

function setData($age=0){ }

this forces the age to be a number, if a string is passed, the value will be 0

for array values , there are two variation

function setData(array $data){ } 

if an array is not passed, it would throw an error

function setData($data=array()){ } 

This would pass an empty array of no value is given for $data

Solution 5

string, int and other built-in types are not classes, in argument you specify class, of the argument. The only supported built-in type to be put there is array.

View more solutions

Comments

[Gumbo]

It does know the types, it just doesn’t care.

[zerobandwidth]

Specifying default values for arguments is an important technique but shouldn't be equated with type hinting. For example, it's not entirely true that specifying func( $x=0 ) forces the value of $x to be a number. Rather, if your function always processes the value of $x as a number throughout, then by coincidence it works out. Actually if you pass in a string that happens to be parsable as a number, then it will be processed that way. func( '5' ) would cheerfully use 5 as the integer value. A string that isn't parsable is interpreted as zero, so your 0 default works by coincidence.

[phil294]

this answer is outdated, see php.net/manual/en/…

[ankit suthar]

Please add more description regarding your answer.

[Roger Campanera]

You can use array as hint since PHP 5.1.0

[Machado]

Of course, you're right, I wrote it right myself after writing wrong hehe.

[Alex Lacayo]

is it always recommended to include parameter type?

[a-ctor]

boolean as type declaration will check for instanceof boolean instead of checking for a boolean value. The correct type declaration is bool. e.g. function a(value: boolean); a(true) will fail.

[johannchopin]

What about the syntax of the return type of this function?

[arm]

What if I have a class Person and I want to accept an array of Person as parameter for my function? How can I do that?

[Daniel Wu]

@AlexLacayo I think it's good practice. PHP has always been very lenient when it comes to variable type and assigment. But being too open will inevitably lead to head scratching bug one day. Return type is declared using a colon, followed by return type. For example : function foo() : int { return 0; }

[mehov]

Shouldn't it be bool instead of boolean according to php.net/manual/en/language.types.declarations.php?

[Amal Ajith]

@mehov yes, type support have become more robust in PHP since I answered it in 2017 when the documentation had boolean instead of bool. I will update the answer so it helps others who see it now.