Question

How to use strict types with default parameter values in PHP

Answer 1

declare(strict_types=1);

function greet(string $name = "Guest"): string {
    return "Hello, $name";
}

echo greet() . "\n";           // OK
echo greet("Sam Sung") . "\n"; // OK
echo greet(123);        // TypeError



/*
run:

ERROR!
Hello, Guest
Hello, Sam Sung

Fatal error: Uncaught TypeError: greet(): Argument #1 ($name) must be of type string, int given

*/

 

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