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.