Declare strict_types directive in PHP - Step-by-Step Execution

Concept Flow - Declare strict_types directive

Start PHP file

↓

Declare strict_types=1?

No→Default weak typing

Yes↓

Enable strict type checking

↓

Run code with strict types

↓

Type errors if mismatch

↓

End

The strict_types directive tells PHP to enforce exact types for function calls, stopping automatic type conversion.

Execution Sample

PHP

<?php
declare(strict_types=1);
function add(int $a, int $b): int {
  return $a + $b;
}
echo add(1, 2);

This code enables strict types and defines a function that adds two integers, then prints the result.

Execution Table

Step	Action	Evaluation	Result
1	Start script with declare(strict_types=1);	strict_types enabled	Strict mode ON
2	Call add(1, 2)	Check argument types	Both are int, OK
3	Execute add function	Return 1 + 2	3
4	Print result	Output 3	3
5	Call add('1', 2)	Check argument types	TypeError thrown (string not int)

💡 Execution stops with TypeError if argument types do not exactly match when strict_types=1

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 5
$a	undefined	1	1	TypeError (no value)
$b	undefined	2	2	TypeError (no value)
Return value	undefined	undefined	3	No return due to error

Key Moments - 2 Insights

Why does add('1', 2) cause an error with strict_types=1?

Because strict_types=1 requires exact types. The string '1' is not automatically converted to int, so PHP throws a TypeError as shown in execution_table step 5.

What happens if declare(strict_types=1) is missing?

PHP uses weak typing and automatically converts types. So add('1', 2) would work by converting '1' to int 1, no error occurs.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output at step 4?

A 3

B TypeError

C 1

D No output

💡 Hint

Check the 'Result' column at step 4 in execution_table

At which step does the TypeError occur when calling add('1', 2)?

A Step 2

B Step 5

C Step 3

D Step 4

💡 Hint

Look for the row mentioning TypeError in execution_table

If declare(strict_types=1) was removed, what would happen at step 5?

A Script stops immediately

B TypeError still thrown

C Function call succeeds with automatic type conversion

D Output is empty

💡 Hint

Refer to key_moments explanation about behavior without strict_types

Concept Snapshot

declare(strict_types=1);
- Placed at top of PHP file
- Enables strict type checking for function calls
- Requires exact argument types (no automatic conversion)
- Causes TypeError if types mismatch
- Helps catch bugs early by enforcing types

Full Transcript

This visual trace shows how the declare(strict_types=1) directive in PHP enforces strict type checking. When enabled, PHP requires function arguments to match declared types exactly. The example code defines a function add(int $a, int $b) and calls it with integers 1 and 2, which works fine and outputs 3. However, calling add('1', 2) causes a TypeError because the string '1' is not automatically converted to int under strict types. The variable tracker shows how $a and $b get values or cause errors. Key moments clarify why strict types cause errors and what happens if the directive is missing. The quiz tests understanding of output, error steps, and behavior changes without strict types.

Practice

1. What does the declare(strict_types=1); directive do in a PHP file?

easy

A. It enables automatic type conversion for function arguments.

B. It disables all error reporting in the PHP script.

C. It sets the default timezone for the script.

D. It forces PHP to check that function arguments match their declared types exactly.

Solution

1. Step 1: Understand the purpose of strict_types

   The directive declare(strict_types=1); tells PHP to enforce exact type matching for function arguments and return values.

2. Step 2: Compare with other options

   Options B, C, and D describe unrelated PHP features: error reporting, timezone setting, and type conversion, which are not controlled by this directive.

3. Final Answer:

   It forces PHP to check that function arguments match their declared types exactly. -> Option D

4. Quick Check:

   strict_types enforces exact type checks = A [OK]

Hint: Strict types means exact argument types only [OK]

Common Mistakes:

• Thinking it disables errors
• Confusing with timezone settings
• Assuming it enables type coercion

2. Which of the following is the correct way to enable strict types in a PHP file?

easy

A.

B.

C.

D.

Solution

1. Step 1: Check correct syntax for declare directive

   The correct syntax is declare(strict_types=1); placed at the top of the PHP file.

2. Step 2: Analyze each option

   <?php declare(strict_types=1); ?> uses the exact correct syntax. Options A and B use invalid PHP syntax. <?php declare(strict_types = true); ?> uses true instead of 1, which is invalid for this directive.

3. Final Answer:

   <?php declare(strict_types=1); ?> -> Option C

4. Quick Check:

   declare(strict_types=1) is correct syntax = C [OK]

Hint: Use declare(strict_types=1); at file top [OK]

Common Mistakes:

• Using true instead of 1
• Trying to assign strict_types as a variable
• Calling a non-existent function

3. Given this PHP code with strict types enabled:

<?php
declare(strict_types=1);
function add(int $a, int $b): int {
    return $a + $b;
}
echo add(5, '3');
?>

What will be the output?

medium

A. TypeError: Argument 2 passed must be of the type int, string given

B. 8

C. 53

D. Warning: Implicit conversion from string to int

Solution

1. Step 1: Understand strict_types effect on function calls

   With declare(strict_types=1);, PHP requires exact types. Passing a string '3' to a parameter typed as int causes an error.

2. Step 2: Predict the runtime behavior

   PHP throws a TypeError because the second argument is a string, not an int, and no automatic conversion happens.

3. Final Answer:

   TypeError: Argument 2 passed must be of the type int, string given -> Option A

4. Quick Check:

   strict_types=1 + wrong argument type = TypeError [OK]

Hint: Strict types cause TypeError on wrong argument types [OK]

Common Mistakes:

• Assuming PHP converts string '3' to int automatically
• Expecting output 8 or 53
• Confusing warning with error

4. Identify the error in this PHP code snippet:

<?php
function multiply(int $x, int $y): int {
    return $x * $y;
}

declare(strict_types=1);
echo multiply(4, 5);
?>

medium

A. The declare(strict_types=1); must be at the top before any code.

B. The function multiply must return a string, not int.

C. The function parameters should not have types declared.

D. There is no error; the code runs fine.

Solution

1. Step 1: Check position of declare(strict_types=1);

   The strict_types directive must be declared at the very top of the PHP file, before any code or whitespace.

2. Step 2: Analyze the code order

   Here, the function is declared before the strict_types directive, which is invalid and causes the directive to be ignored or cause an error.

3. Final Answer:

   The declare(strict_types=1); must be at the top before any code. -> Option A

4. Quick Check:

   declare(strict_types=1) must be first line = B [OK]

Hint: Put declare(strict_types=1); at file very top [OK]

Common Mistakes:

• Placing declare after code or functions
• Thinking return type must be string
• Removing type declarations from parameters

5. You want to write a PHP function that accepts only integers and returns their sum. You also want to ensure no accidental type coercion happens anywhere in the file. How should you set up your PHP file?

hard

A. Use no type declarations and rely on PHP's automatic type conversion.

B. Place declare(strict_types=1); at the top, then define the function with int types.

C. Use declare(strict_types=0); and define the function with int types.

D. Define the function with int types but do not use declare(strict_types=1);

Solution

1. Step 1: Enable strict types at the file start

   To prevent type coercion, add declare(strict_types=1); at the very top of the PHP file.

2. Step 2: Define the function with int type hints

   Declare the function parameters and return type as int to enforce integer-only usage.

3. Final Answer:

   Place declare(strict_types=1); at the top, then define the function with int types. -> Option B

4. Quick Check:

   strict_types=1 + int types = strict integer enforcement [OK]

Hint: Use declare(strict_types=1); plus int type hints [OK]

Common Mistakes:

• Skipping declare(strict_types=1);
• Using declare(strict_types=0); which allows coercion
• Not adding type hints in function