📖 Scenario: Imagine you are building a simple system to manage different types of workers in a company. Some workers can both write reports and attend meetings. You want to make sure that certain functions only accept workers who can do both tasks.
🎯 Goal: You will create interfaces for ReportWriter and MeetingAttendee, then create a class that implements both. Finally, you will write a function that accepts only objects that satisfy both interfaces using intersection types.
📋 What You'll Learn
Create two interfaces: ReportWriter and MeetingAttendee with one method each.
Create a class Employee that implements both interfaces.
Write a function handleWorker that accepts a parameter with an intersection type of ReportWriter&MeetingAttendee.
Call the function with an Employee object and print messages from both methods.
💡 Why This Matters
🌍 Real World
Intersection types help ensure that objects passed to functions meet multiple requirements, useful in complex systems where roles overlap.
💼 Career
Understanding intersection types is important for writing clear, type-safe code in PHP, especially in large projects or when using modern PHP features.
Progress0 / 4 steps
1
Create interfaces for report writing and meeting attendance
Create two interfaces called ReportWriter and MeetingAttendee. In ReportWriter, declare a public method writeReport(). In MeetingAttendee, declare a public method attendMeeting().
PHP
Hint
Use the interface keyword to create interfaces. Each interface should have one method declaration without a body.
2
Create a class that implements both interfaces
Create a class called Employee that implements both ReportWriter and MeetingAttendee. Implement the methods writeReport() and attendMeeting() to return the strings "Report written." and "Meeting attended." respectively.
PHP
Hint
Use implements keyword to implement multiple interfaces. Each method must return the exact string.
3
Write a function using intersection types
Write a function called handleWorker that accepts one parameter with the intersection type ReportWriter&MeetingAttendee. Inside the function, call writeReport() and attendMeeting() on the parameter and store their results in variables $report and $meeting.
PHP
Hint
Use the intersection type ReportWriter&MeetingAttendee in the function parameter to require both interfaces.
4
Call the function and print the results
Inside the handleWorker function, add two echo statements to print $report and $meeting each followed by a newline. Then, create an Employee object called $employee and call handleWorker($employee).
PHP
Hint
Use echo to print the strings with a newline character \n. Create the object and call the function after it.
Practice
(1/5)
1. What does an intersection type in PHP represent? function example(SomeType&OtherType $param) {}
easy
A. A type that requires the value to satisfy both SomeType and OtherType.
B. A type that accepts either SomeType or OtherType.
C. A type that only accepts SomeType but ignores OtherType.
D. A type that combines SomeType and OtherType into a new class.
Solution
Step 1: Understand intersection type syntax
In PHP, the & symbol between types means the value must satisfy all listed types simultaneously.
Step 2: Interpret the function parameter
The parameter type SomeType&OtherType means the argument must be both SomeType and OtherType at the same time.
Final Answer:
A type that requires the value to satisfy both SomeType and OtherType. -> Option A
Quick Check:
Intersection type = must satisfy all types [OK]
Hint: Remember & means AND, so all types must match [OK]
Common Mistakes:
Confusing & with | (union type)
Thinking it creates a new combined class
Assuming it accepts either type instead of both
2. Which of the following is the correct syntax for declaring an intersection type in a function parameter in PHP?
easy
A. function test(A & B $param) {}
B. function test(A | B $param) {}
C. function test(A, B $param) {}
D. function test(A && B $param) {}
Solution
Step 1: Recall intersection type syntax
PHP uses a single ampersand & to declare intersection types between multiple types.
Step 2: Check each option
function test(A & B $param) {} uses & correctly. function test(A | B $param) {} uses | which is union type. function test(A, B $param) {} uses comma which is invalid. function test(A && B $param) {} uses && which is not valid syntax.
Final Answer:
function test(A & B $param) {} -> Option A
Quick Check:
Intersection type uses & symbol [OK]
Hint: Use single & between types for intersection [OK]
Common Mistakes:
Using | instead of &
Using commas to separate types
Using && instead of &
3. What will be the output of the following PHP code?
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. TypeError at runtime
B. foo
C. bar
D. foo bar
Solution
Step 1: Understand class and interfaces
Class C implements both interfaces A and B, so it has methods foo() and bar().
Step 2: Analyze function test parameter and call
The function test requires an argument that satisfies both A and B (intersection type). Object $c satisfies both. The function echoes foo() and bar() results concatenated with a space.
Final Answer:
foo bar -> Option D
Quick Check:
Object satisfies both interfaces, prints both methods [OK]
Hint: Intersection means object must have all methods used [OK]
Common Mistakes:
Expecting only one method output
Thinking it causes a TypeError
Confusing intersection with union types
4. Identify the error in this PHP code using intersection types:
interface X { public function x(): void; } interface Y { public function y(): void; }
class Z implements X { public function x(): void {} }
function foo(X & Y $param) { $param->x(); $param->y(); }
$z = new Z(); foo($z);
medium
A. Function foo has wrong parameter syntax for intersection types.
B. Method y() is missing in function foo.
C. Class Z does not implement interface Y, so it cannot satisfy X & Y.
D. No error, code runs fine.
Solution
Step 1: Check class Z implementation
Class Z implements only interface X, not Y, so it does not satisfy the intersection type X & Y.
Step 2: Analyze function call with $z
Function foo requires a parameter that implements both X and Y. Passing $z causes a TypeError because $z lacks Y.
Final Answer:
Class Z does not implement interface Y, so it cannot satisfy X & Y. -> Option C
Quick Check:
Intersection requires all interfaces implemented [OK]
Hint: Check all interfaces are implemented for intersection [OK]
Common Mistakes:
Assuming partial interface implementation is enough
Thinking syntax is wrong instead of logic
Ignoring missing methods from interfaces
5. Given these interfaces and classes:
interface Logger { public function log(string $msg): void; } interface FileWriter { public function write(string $data): void; }
class FileLogger implements Logger, FileWriter { public function log(string $msg): void { echo "Log: $msg\n"; } public function write(string $data): void { echo "Write: $data\n"; } }