return.missing

Code example

<?php declare(strict_types = 1);

function getGreeting(string $name): string
{
	$greeting = 'Hello, ' . $name;
}

Why is it reported?

A function or method declares a non-void return type but does not always return a value. Every code path must end with a return statement when the function has a return type. Falling off the end of the function without returning a value causes PHP to return null, which violates the declared return type.

When the function has a native return type, this error is not ignorable because PHP will throw a TypeError at runtime.

How to fix it

Add the missing return statement:

 function getGreeting(string $name): string
 {
 	$greeting = 'Hello, ' . $name;
+	return $greeting;
 }

Or change the return type if the function is not meant to return a value:

-function getGreeting(string $name): string
+function getGreeting(string $name): void
 {
 	$greeting = 'Hello, ' . $name;
+	echo $greeting;
 }