Type erasure

Generic type arguments exist only at compile time. The compiler checks them, then erases them to their bounds and inserts casts, so the bytecode carries no record of the parameterisation.

List<String> and List<Integer> compile to the same class. The angle brackets are a proof obligation discharged at compile time, not a runtime property.

Consequences you actually hit

new T[10] does not compile — the array’s runtime component type is unknown.
instanceof List<String> does not compile; only instanceof List<?> does.
Two methods differing only in their generic parameter clash after erasure.

void f(List<String> a) {}
void f(List<Integer> a) {}   // name clash: same erasure

The escape hatch

Erasure is not total: generic signatures are retained in a separate class-file attribute for reflection and for the compiler reading a library. Class.getGenericSuperclass() reads it, which is how the super-type-token trick works.

abstract class TypeRef<T> {
    final Type type = ((ParameterizedType) getClass()
        .getGenericSuperclass()).getActualTypeArguments()[0];
}
Type t = new TypeRef<List<String>>() {}.type;   // List<String>

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