Java type erase - why can I see the type when I look at bytecode?

Question

I'm trying to understand why writing both methods in a class is not allowed

 public bool plus(List<String>) {return true;}
 public bool plus(List<Integer>) {return true;}

I am trying to understand how this is related to type-erasing, but when I decompile the following code

public class Test<T> {

   boolean plus2(List<T> ss) {return false;}
   boolean plus(List<String> ss) {return false;}
   boolean plus(Set<Integer> ss) {return false;}
}

I get the same when I decompile it using Java decompiler (jd)

Even when I print the byte code, I can clearly see the types.
(Looking at the SO answer that declares 'but be sure types are erased in bytecode')

    Compiled from "Test.java"
    public class com.example.Test<T> {
    public com.example.Test();
            Code:
            0: aload_0
            1: invokespecial #1                  // Method java/lang/Object."<init>":()V
            4: return

            boolean plus2(java.util.List<T>);
            Code:
            0: iconst_0
            1: ireturn

            boolean plus(java.util.List<java.lang.String>);
            Code:
            0: iconst_0
            1: ireturn

            boolean plus(java.util.Set<java.lang.Integer>);
            Code:
            0: iconst_0
            1: ireturn
            }

Answer 1

Your compiler should be able to verify information of a general type based on information in byte code.

Java 5.0+ , .

.

 // list.getClass() is just ArrayList which doesn't record the generic type.
List list = new ArrayList<String>();

// list.getClass() is not ArrayList
// list is an instance of a class which extends ArrayList<String>
List list = new ArrayList<String>() { };

ParameterizedType t = (ParameterizedType) list.getClass().getGenericSuperclass();
assert t.getRawType() == ArrayList.class;
assert t.getActualTypeArguments()[0] == String.class;

, - ArrayList , .

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