parameters in class data type function (php)

Question

when I see parameters in a function in the form of a class name, what does it mean? how can you google it? it is clear that if there is an array in front of a variable, it must be a variable of the array type, but if the class is in front of the variable, what does this mean?

public function scopeFilter(Builder $builder, array $filter)
     { /**/ }

I tried to google but couldn't find it

Builder is object of class, array is array if there is no indicated type - then you need to read the function code and deduce what it can be. — WiatroBosy
– WiatroBosy, Commented Dec 31, 2022 at 14:48

Peppermintology · Accepted Answer · 2022-12-31 14:50:33Z

1

It is known as Type hinting and has been part of the PHP core since version 5.

As per the documentation:

They ensure that the value is of the specified type at call time, otherwise a TypeError is thrown.

So using your example, if you tried to call your scopeFilter function and provide it with arguments that were not a Builder and an array, you'd get a runtime error.

There are plenty of articles available should you want to read more about type hinting.

answered Dec 31, 2022 at 14:50

Peppermintology

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