Declare strict_types directive in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output with strict_types enabled

What is the output of this PHP code when declare(strict_types=1); is used?

PHP

<?php

declare(strict_types=1);

function addInts(int $a, int $b): int {
    return $a + $b;
}

try {
    echo addInts(5, '10');
} catch (TypeError $e) {
    echo 'TypeError caught';
}

A 15

B TypeError caught

C 5

D 10

💡 Hint

With strict_types=1, PHP does not convert types automatically for function arguments.

When strict_types is enabled, passing a string '10' to a function expecting int causes a TypeError. Without strict_types, PHP would convert '10' to int 10.

❓ Predict Output

intermediate

Behavior without strict_types directive

What will this PHP code output when declare(strict_types=1); is NOT present?

PHP

<?php

function multiplyInts(int $x, int $y): int {
    return $x * $y;
}

echo multiplyInts(3, '4');

A 12

B TypeError caught

C 7

D 34

💡 Hint

Without strict_types, PHP converts scalar types automatically.

Without strict_types, PHP converts the string '4' to integer 4 automatically, so the multiplication works and outputs 12.

❓ Predict Output

advanced

Return type enforcement with strict_types

What is the output of this PHP code with declare(strict_types=1);?

PHP

<?php

declare(strict_types=1);

function getNumber(): int {
    return '5';
}

try {
    echo getNumber();
} catch (TypeError $e) {
    echo 'Return type error';
}

A 0

B TypeError caught

C 5

D Return type error

💡 Hint

Strict types also enforce return types, not just argument types.

With strict_types=1, returning a string '5' from a function declared to return int causes a TypeError at runtime.

❓ Predict Output

advanced

Effect of strict_types on scalar type coercion in method calls

Given this PHP code with declare(strict_types=1);, what will be the output?

PHP

<?php

declare(strict_types=1);

class Calculator {
    public function subtract(int $a, int $b): int {
        return $a - $b;
    }
}

$calc = new Calculator();
try {
    echo $calc->subtract(10, '3');
} catch (TypeError $e) {
    echo 'TypeError caught';
}

A TypeError caught

B 103

C 7

D 13

💡 Hint

Method argument types are also strictly enforced with strict_types=1.

Passing a string '3' to a method expecting int with strict_types=1 causes a TypeError.

🧠 Conceptual

expert

Understanding strict_types directive scope

Which statement about the declare(strict_types=1); directive in PHP is TRUE?

A It disables all type checking for scalar types in the file where declared.

B It enforces strict typing globally across all included files regardless of their own declarations.

C It only affects the file where it is declared and does not affect included or required files.

D It automatically converts all scalar types to the expected type without error.

💡 Hint

Think about how PHP treats strict_types in separate files.

The strict_types directive applies only to the file where it is declared. Included files must declare it separately to enforce strict typing.