Why is dynamic assignment required in a particular case in Byte Buddy? - Stack Overflow

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I have a TypeDescription.Generic built like this:

import static net.bytebuddy.description.type.TypeDescription.Generic.Builder.parameterizedType;

final TypeDescription.Generic supplierType =
  parameterizedType(typePool.describe("java.util.function.Supplier").resolve(),
                    List.of(TypeDescription.Generic.Builder.of(mySuperclass.asGenericType()).asWildcardUpperBound()))
      .build();

(mySuperclass is a non-generic class, i.e. a simple TypeDescription.)

I believe that supplierType accurately represents java.util.function.Supplier<? extends MySuperclass>.

Next, in a DynamicType.Builder (using a subclassing strategy), I define a private field using this supplierType:

builder = builder.defineField("$supplier", supplierType, PRIVATE, SYNTHETIC, FieldManifestation.FINAL);

I believe that this field definition accurately represents private final Supplier<? extends MySuperclass> $supplier;.

Next, I define a method returning the result of invoking get() on that Supplier:

builder = builder
  .defineMethod("$get", mySuperclass, PUBLIC, SYNTHETIC, MethodManifestation.FINAL)
  .intercept(MethodCall.invoke(named("get"))
             .onField("$supplier")
             .withAssigner(Assigner.DEFAULT, 
                           Assigner.Typing.DYNAMIC)); // <-- my question

I believe that this method definition accurately represents public final MySuperclass $get() { return $supplier.get(); }, but I have a question about its return type and casting and dynamic assignment.

Without the withAssigner configuration, at runtime the $get() method in my generated class fails, saying basically that the Supplier in question cannot return a T as a mySuperclass.

I thought that because of the upper bound of the Supplier's wildcard type argument, which is mySuperclass, I wouldn't need this dynamic assignment. What have I misunderstood?

I have a TypeDescription.Generic built like this:

import static net.bytebuddy.description.type.TypeDescription.Generic.Builder.parameterizedType;

final TypeDescription.Generic supplierType =
  parameterizedType(typePool.describe("java.util.function.Supplier").resolve(),
                    List.of(TypeDescription.Generic.Builder.of(mySuperclass.asGenericType()).asWildcardUpperBound()))
      .build();

(mySuperclass is a non-generic class, i.e. a simple TypeDescription.)

I believe that supplierType accurately represents java.util.function.Supplier<? extends MySuperclass>.

Next, in a DynamicType.Builder (using a subclassing strategy), I define a private field using this supplierType:

builder = builder.defineField("$supplier", supplierType, PRIVATE, SYNTHETIC, FieldManifestation.FINAL);

I believe that this field definition accurately represents private final Supplier<? extends MySuperclass> $supplier;.

Next, I define a method returning the result of invoking get() on that Supplier:

builder = builder
  .defineMethod("$get", mySuperclass, PUBLIC, SYNTHETIC, MethodManifestation.FINAL)
  .intercept(MethodCall.invoke(named("get"))
             .onField("$supplier")
             .withAssigner(Assigner.DEFAULT, 
                           Assigner.Typing.DYNAMIC)); // <-- my question

I believe that this method definition accurately represents public final MySuperclass $get() { return $supplier.get(); }, but I have a question about its return type and casting and dynamic assignment.

Without the withAssigner configuration, at runtime the $get() method in my generated class fails, saying basically that the Supplier in question cannot return a T as a mySuperclass.

I thought that because of the upper bound of the Supplier's wildcard type argument, which is mySuperclass, I wouldn't need this dynamic assignment. What have I misunderstood?

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Share Improve this question Follow edited Feb 4 at 19:13 Laird Nelson asked Feb 4 at 18:47 Laird NelsonLaird Nelson 16.3k2121 gold badges7979 silver badges137137 bronze badges

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1 Answer 1

Sorted by: Reset to default 1

Byte Buddy works on byte code. You do not have to add the cast in Java, because the compiler adds it for you, but it will still be present in the byte code. get will return an object type, and it is then casted to T by the compiler.

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