Java Generics: Type Erasure, Wildcards and PECS

Generics let a class or method work with many types while the compiler still checks them. List<String>, Map<String, Integer> and Optional<User> are all generic types. Interviewers ask about three ideas: type erasure, invariance, and wildcards (PECS).

Why generics exist

Before Java 5, collections held Object, so you had to cast on every read, and a wrong cast failed only at run time with a ClassCastException. Generics move that check to compile time:

import java.util.*;

public class RawTypes {
    @SuppressWarnings({"rawtypes", "unchecked"})
    public static void main(String[] args) {
        List raw = new ArrayList();                  // raw type: avoid
        raw.add("text");
        try {
            Integer wrong = (Integer) raw.get(0);    // compiles, then fails at run time
        } catch (ClassCastException e) {
            System.out.println("ClassCastException at run time");   // ClassCastException at run time
        }

        List<String> safe = new ArrayList<>();
        safe.add("text");
        // safe.add(42);                             // compile error: caught immediately
    }
}

Never use raw types in new code: they switch off the compiler's checks.

Generic methods and bounded types

import java.util.*;

public class GenericsDemo {
    public static void main(String[] args) {
        System.out.println(max(3, 7));                       // 7
        System.out.println(max("apple", "pear"));            // pear

        List<Integer> ints = List.of(1, 2, 3);
        List<Double> doubles = List.of(1.5, 2.5);
        System.out.println(sum(ints) + sum(doubles));        // 10.0

        List<Number> sink = new ArrayList<>();
        addIntegers(sink);                                   // a List<? super Integer> accepts a List<Number>
        System.out.println(sink);                            // [1, 2]

        System.out.println(new ArrayList<String>().getClass() == new ArrayList<Integer>().getClass());   // true  (type erasure)
    }

    static <T extends Comparable<T>> T max(T a, T b) {       // bounded type: T must be comparable
        return a.compareTo(b) >= 0 ? a : b;
    }

    static double sum(List<? extends Number> numbers) {      // producer: we only read from it
        double total = 0;
        for (Number n : numbers) total += n.doubleValue();
        return total;
    }

    static void addIntegers(List<? super Integer> target) {  // consumer: we only write to it
        target.add(1);
        target.add(2);
    }
}

<T extends Comparable<T>> is a bounded type parameter: it lets the method call compareTo on T.

Type erasure

Generics exist only at compile time. The compiler checks the types, inserts casts, and then erases the type arguments, so at run time List<String> and List<Integer> are both just List (the last line above prints true). This was done for backward compatibility with pre-generics code.

Consequences you can be asked about:

You cannot write new T() or new T[10]: the type is unknown at run time.
You cannot use instanceof List<String>. Only instanceof List<?> is allowed.
You cannot overload void f(List<String>) and void f(List<Integer>): after erasure they have the same signature.
Primitives cannot be type arguments (List<int> is illegal), so you use List<Integer> and pay for autoboxing.

Invariance: List<Object> is not a List<String>

Arrays are covariant (String[] is an Object[]), but generics are invariant: List<String> is not a List<Object>, even though String is an Object. If it were allowed, you could add an Integer to what is really a list of strings.

import java.util.*;

public class Invariance {
    public static void main(String[] args) {
        List<String> strings = new ArrayList<>(List.of("a", "b"));
        // List<Object> objects = strings;        // compile error: generics are invariant
        List<? extends Object> readOnly = strings;    // allowed: a read-only view of "some kind of Object"
        System.out.println(readOnly.get(0));          // a

        Object[] array = new String[1];               // arrays ARE covariant...
        try {
            array[0] = 42;                            // ...so this compiles but fails at run time
        } catch (ArrayStoreException e) {
            System.out.println("ArrayStoreException");   // ArrayStoreException
        }
    }
}

Wildcards and PECS

Wildcard	Meaning	You can...
`List<?>`	A list of some unknown type	Read items as `Object`. Add only `null`
`List<? extends Number>`	A list of `Number` or a subtype	Read items as `Number`. You cannot add
`List<? super Integer>`	A list of `Integer` or a supertype	Add `Integer`s. Reads come back as `Object`

PECS: Producer Extends, Consumer Super.

If a collection produces values you read (a source), use ? extends T.
If it consumes values you write (a destination), use ? super T.

That is why Collections.copy(List<? super T> dest, List<? extends T> src) has exactly that signature. In the example above, reads from the list, so it accepts a and a . writes to it, so it accepts a or .

Common interview answers

What is type erasure? The compiler removes generic type information after checking, so it does not exist at run time.
Why can't you create a generic array? The array's element type must be known at run time (arrays are reified), but erasure removes it.
versus ? accepts a list of any type (read-only in practice). accepts only an exact and allows adding any object.

Practise

Write and test it.
Write and call it with a and a .
Explain PECS with the signature.

Java Generics: Type Erasure, Wildcards and PECS

Stage 2: Object-oriented programming

Stage 3: Collections, generics and exceptions

Stage 4: Modern Java and streams

Stage 5: Concurrency and the JVM

Stage 6: Databases, Spring and testing

Stage 7: Interview practice