Intersection types in practice in PHP - Mini Project: Build & Apply

Intersection types in practice

📖 Scenario: Imagine you are building a simple system to manage different types of workers in a company. Some workers can both code and design. You want to make sure that certain functions only accept workers who can do both tasks.

🎯 Goal: You will create interfaces for coding and designing, then create a class that implements both. Finally, you will write a function that accepts only workers who are both coders and designers using intersection types.

📋 What You'll Learn

Create two interfaces: Coder and Designer

Create a class FullStackWorker that implements both Coder and Designer

Write a function workOnProject that accepts a parameter with an intersection type of Coder&Designer

Call the workOnProject function with an instance of FullStackWorker and print the result

💡 Why This Matters

🌍 Real World

Intersection types help ensure that objects passed to functions have multiple capabilities, useful in systems where roles overlap, like full-stack developers who code and design.

💼 Career

Understanding intersection types is important for writing clear, type-safe code in modern PHP applications, improving code quality and reducing bugs.

1

Create interfaces for coding and designing

Create two interfaces called Coder and Designer. Each interface should have one public method: code() for Coder and design() for Designer. Both methods should return a string.

PHP

<?php

// Create interface Coder with method code() returning string
// Create interface Designer with method design() returning string

Hint

Use the interface keyword to create interfaces. Define methods with public visibility and specify the return type string.

2

Create a class that implements both interfaces

Create a class called FullStackWorker that implements both Coder and Designer interfaces. Implement the code() method to return the string 'Coding...' and the design() method to return the string 'Designing...'.

PHP

<?php

interface Coder {
    public function code(): string;
}

interface Designer {
    public function design(): string;
}

// Create class FullStackWorker implementing Coder and Designer
// Implement code() to return 'Coding...'
// Implement design() to return 'Designing...'

Hint

Use implements keyword to implement multiple interfaces. Define both methods exactly as in the interfaces.

3

Write a function using intersection types

Write a function called workOnProject that accepts one parameter called worker with the intersection type Coder&Designer. Inside the function, return the concatenation of $worker->code() and $worker->design() separated by a space.

PHP

code() . ' ' . $worker->design()

Hint

Use the intersection type Coder&Designer in the function parameter to require both interfaces.

4

Call the function and print the result

Create an instance of FullStackWorker called worker. Call the function workOnProject with worker as argument and print the returned string.

PHP

code() . ' ' . $worker->design();
}

// Create instance worker of FullStackWorker
// Print the result of workOnProject(worker)

Hint

Create the object with new FullStackWorker() and use print() to show the result.

Practice

1. What does an intersection type like A&B mean in PHP?

easy

A. The value must be type B only.

B. The value can be either type A or type B.

C. The value must satisfy both type A and type B simultaneously.

D. The value must be type A only.

Solution

1. Step 1: Understand intersection type concept

   Intersection types require a value to meet all listed types at once, not just one.

2. Step 2: Apply to A&B

   The value must be both type A and type B simultaneously.

3. Final Answer:

   The value must satisfy both type A and type B simultaneously. -> Option C

4. Quick Check:

   Intersection means AND, so both types required [OK]

Hint: Intersection means all types must be met together [OK]

Common Mistakes:

• Confusing intersection (&) with union (|)
• Thinking it means either type A or B
• Assuming it means only one type

2. Which of the following is the correct syntax for declaring a function parameter with an intersection type of interfaces Countable and Iterator?

easy

A. function process(Countable&Iterator $obj) {}

B. function process(Countable|Iterator $obj) {}

C. function process(Countable, Iterator $obj) {}

D. function process(Countable & Iterator $obj) {}

Solution

1. Step 1: Recall intersection type syntax in PHP

   PHP uses the ampersand (&) without spaces and escaped as & in HTML to declare intersection types.

2. Step 2: Check each option

   function process(Countable&Iterator $obj) {} uses Countable&Iterator which is correct syntax for intersection types in PHP.

3. Final Answer:

   function process(Countable&Iterator $obj) {} -> Option A

4. Quick Check:

   Intersection uses & without spaces [OK]

Hint: Use & with no spaces for intersection types [OK]

Common Mistakes:

• Using | instead of & for intersection
• Adding spaces around &
• Using comma instead of &

3. Given the following code, what will be the output?

interface A { public function foo(): string; }
interface B { public function bar(): string; }

class C implements A, B {
  public function foo(): string { return 'foo'; }
  public function bar(): string { return 'bar'; }
}

function test(A&B $obj) {
  echo $obj->foo() . ' ' . $obj->bar();
}

$c = new C();
test($c);

medium

A. foo bar

B. foo

C. bar

D. TypeError at runtime

Solution

1. Step 1: Understand class C implements both interfaces

   Class C implements both A and B, so it has methods foo() and bar().

2. Step 2: Call test with intersection type parameter

   Function test requires an object that is both A and B, which $c is. It calls foo() and bar() and echoes their results separated by space.

3. Final Answer:

   foo bar -> Option A

4. Quick Check:

   Object meets both interfaces, so output is 'foo bar' [OK]

Hint: Check if object implements all interfaces in intersection [OK]

Common Mistakes:

• Expecting only one method output
• Thinking intersection causes error if both interfaces implemented
• Confusing union with intersection types

4. What is the error in this code snippet?

interface X { public function x(): void; }
interface Y { public function y(): void; }

function example(X&Y $obj) {
  $obj->x();
  $obj->y();
}

class Z implements X {
  public function x(): void {}
}

$z = new Z();
example($z);

medium

A. No error, code runs fine.

B. TypeError because $z does not implement interface Y.

C. Syntax error in function declaration.

D. Fatal error: method y() not found in class Z.

Solution

1. Step 1: Check class Z interfaces

   Class Z implements only interface X, not Y.

2. Step 2: Analyze function parameter type

   Function example requires an object implementing both X and Y (intersection). Passing $z violates this.

3. Final Answer:

   TypeError because $z does not implement interface Y. -> Option B

4. Quick Check:

   Intersection requires all interfaces, missing one causes TypeError [OK]

Hint: All interfaces in intersection must be implemented [OK]

Common Mistakes:

• Assuming partial interface implementation is enough
• Confusing syntax error with type error
• Thinking missing method causes fatal error instead of TypeError

5. You want to write a function that accepts an object which must be both JsonSerializable and Countable. You also want to ensure the object has a method customMethod(). Which of the following is the best way to type hint this in PHP 8.1+?

hard

A. function process(JsonSerializable|Countable $obj) { $obj->customMethod(); }

B. function process(JsonSerializable&Countable&CustomInterface $obj) {}

C. function process(object $obj) { $obj->customMethod(); }

D. function process(JsonSerializable&Countable $obj) { if (method_exists($obj, 'customMethod')) { $obj->customMethod(); } }

Solution

1. Step 1: Understand intersection type usage

   Use & to require both JsonSerializable and Countable interfaces.

2. Step 2: Handle custom method not in interfaces

   Since customMethod() is not part of these interfaces, check its existence with method_exists before calling.

3. Final Answer:

   function process(JsonSerializable&Countable $obj) { if (method_exists($obj, 'customMethod')) { $obj->customMethod(); } } -> Option D

4. Quick Check:

   Intersection + runtime check for extra method [OK]

Hint: Use & for interfaces and method_exists for extra methods [OK]

Common Mistakes:

• Using union (|) instead of intersection (&)
• Assuming method in interfaces without checking
• Ignoring type safety by using object without interfaces