英文:
Why builder lost type?
问题
以下是翻译好的内容：
当我尝试这样做时：public class Example<T> {private T value;private Example() {}public T getValue() {return value;}public void setValue(T value) {this.value = value;}public static <T> Builder<T> builder() {return new Builder<>();}public static class Builder<T> {public Example<T> build() {return new Example<>();}}public static void main(String[] args) {Example<String> example = Example.builder().build();example.setValue("test");System.out.println(example.getValue().toUpperCase());}}我得到一个错误：Example.java:27: 错误：不兼容的类型：无法将 Example<Object> 转换为 Example<String>Example<String> example = Example.builder().build();但是当我在 builder 方法的参数中添加类型时，一切正常运作：public static <T> Builder<T> builder(T type) {return new Builder<>();}为什么？我如何消除这个虚拟的类型参数？附加问题：如何将复杂类型传递给像这样的方法？public static <T> Builder<T> builder(Class<T> type) {return new Builder<>();}Example<Optional<String>> example = Example.builder(???).build();**更新：** 这只是一个示例。实际应用中将会有多个构建器在构建过程的不同阶段相互创建，并且当然，它们将具有接受泛型类型的方法。我希望有一种方法可以帮助编译器推断类型。我甚至同意使用反射的技巧，如果有的话，它们可以帮助我获得一个漂亮的 DSL。
英文:
When I try something like this
public class Example<T> {private T value;private Example() {}public T getValue() {return value;}public void setValue(T value) {this.value = value;}public static <T> Builder<T> builder() {return new Builder<>();}public static class Builder<T> {public Example<T> build() {return new Example<>();}}public static void main(String[] args) {Example<String> example = Example.builder().build();example.setValue("test");System.out.println(example.getValue().toUpperCase());}}
I get an error
Example.java:27: error: incompatible types: Example<Object> cannot be converted to Example<String>Example<String> example = Example.builder().build();
But when I add type to arguments of builder method, everything works fine
public static <T> Builder<T> builder(T type) {return new Builder<>();}
Why? And how I can rid off this dummy type argument?
Bonus question: How to pass complex type to the method like this?
public static <T> Builder<T> builder(Class<T> type) {return new Builder<>();}Example<Optional<String>> example = Example.builder(???).build();
UPDATE: It's only an example. The real-life application will have several builders creating each other on different stages of the building process and of course, they will have methods accepting generic type. I hope there is a way to help the compiler infer types. I even agree to tricks with reflection, if any exist and they can help me to get a beautiful DSL.
答案1
得分: 2
"Example"是一个原始类，因此隐式地，T被视为Object。您可以在调用方法之前使用尖括号（<>）将泛型类型指定给方法：
Example<String> example = Example.<String>builder().build();// 这里 --------------------------^
英文:
Example is a raw class, so implicitly T is Object. You can specify the generic type when calling the method by putting it in angle-brackets (<>) before the method call:
Example<String> example = Example.<String>builder().build();// Here --------------------------^
答案2
得分: 1
编译器可以推断出您要创建的Example是一个Example<String>，但它无法推断出builder()方法返回的Builder应该是一个Builder<String>。
您可以通过将语句分为两个语句来帮助编译器：
Builder<String> builder = Example.builder();Example<String> example = builder.build();
或者通过指定您从builder()方法中期望的Builder类型来帮助编译器：
Example<String> example = Example.<String>builder().build();
英文:
The compiler can infer that the Example you are trying to create is an Example<String>, but it can't infer that the Builder returned by builder() method should be a Builder<String>.
You can help the compiler by breaking the statement into two statements:
Builder<String> builder = Example.builder();Example<String> example = builder.build();
Or by specifying the type of Builder you expect from the builder() method:
Example<String> example = Example.<String>builder().build();
答案3
得分: 1
只要您不介意未检查的赋值警告，您可以仅进行以下更改：
public static Builder builder() {return new Builder<>();}
编辑开始：
此外，如果您想要省略警告，可以将Builder类更改为：
public static class Builder {private <T> Example<T> build() {return new Example<>();}}
并将静态的Builder方法更改为：
public static Builder builder() {return new Builder();}
编辑结束。
然后您可以像这样使用它：
Example<String> example = Example.builder().build();
英文:
if you don't mind an Unchecked assignment warning you could add just the following change:
public static Builder builder() {return new Builder<>();}
EDIT Start:
Additionally, if you want to omit the warning you can change the Builder class into:
public static class Builder {private <T> Example<T> build() {return new Example<>();}}
and change the static Builder method to:
public static Builder builder() {return new Builder();}
EDIT End.
and then you can use it like this:
Example<String> example = Example.builder().build();
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