Jason S Jason S - 2 months ago 25
Java Question

Java generics + Builder pattern

How do I call

start()
below?

package com.example.test;

class Bar {}

public class Foo<K>
{
final private int count;
final private K key;

Foo(Builder<K> b)
{
this.count = b.count;
this.key = b.key;
}

public static class Builder<K2>
{
int count;
K2 key;

private Builder() {}
static public <K3> Builder<K3> start() { return new Builder<K3>(); }
public Builder<K2> setCount(int count) { this.count = count; return this; }
public Builder<K2> setKey(K2 key) { this.key = key; return this; }
public Foo<K2> build() { return new Foo(this); }
}

public static void main(String[] args)
{
Bar bar = new Bar();
Foo<Bar> foo1 = Foo.Builder.start().setCount(1).setKey(bar).build();
// Type mismatch: cannot convert from Foo<Object> to Foo<Bar>

Foo<Bar> foo2 = Foo.Builder<Bar>.start().setCount(1).setKey(bar).build();
// Multiple markers at this line
// - Bar cannot be resolved
// - Foo.Builder cannot be resolved
// - Syntax error on token ".", delete this token
// - The method start() is undefined for the type Foo<K>
// - Duplicate local variable fooType mismatch: cannot convert from Foo<Object> to Foo<Bar>

Foo<Bar> foo3 = Foo<Bar>.Builder.start().setCount(1).setKey(bar).build();
// Multiple markers at this line
// - Foo cannot be resolved
// - Syntax error on token ".", delete this token
// - Bar cannot be resolved
}
}

Answer

You were close:

Foo.Builder.<Bar> start().setCount(1).setKey(bar).build();

Cheers! :)

P.S. If the compiler can't infer the type parameter of the method on its own, you can force it by calling obj.<Type> method(...) .

P.P.S you might want to use:

public Foo<K2> build() {
    return new Foo<K2>(this);
}

Avoid using raw types.