Question

How to use strict types in PHP

Answer 1

declare(strict_types = 1);

function show(string $name, int $age){
    echo "Name: $name, Age: $age";
}

show("Tom", 52);

// show("Tom", "52"); // error:  Uncaught TypeError: show(): Argument #2 ($age) must be of type int, string given,




/*
run:

Name: Tom, Age: 52

*/

 

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