带有列表的原始类型< String>给编译错误 [英] Raw Type with List<String> gives compilation error

问题描述

我有以下通用类：

  import java.util.ArrayList; 
 import java.util.List; 
 
公共类GenericRaw< T> {
 public List< String> get（）{
返回新的ArrayList<>（）; 
} 
} 
  

让我们考虑一下它的用法： p>

  public class用法{
 public void doSomething（）{
 GenericRaw base = new GenericRaw（）; 
 for（String x：base.get（））{} 
} 
} 
  

对于这段代码，Idea并没有给出任何编译错误，但是Java编译器本身并没有这样做：

java：不兼容类型
required：java.lang.String
found：java.lang.Object

可在JDK上重现1.6.0_33以及JDK 1.7.0_17。

有人可以帮我解释这个问题吗？

我的调查结果。
以下变体可以成功编译：

  public void doSomething（）{
 GenericRaw<> base = new GenericRaw（）; 
 for（String x：base.get（））{} 
} 
  

甚至：

  public void doSomething（）{
 GenericRaw base = new GenericRaw（）; 
列表< String> list = base.get（）; 
 for（String x：list）{} 
} 
  

$

当然。它遵循JLS对原始类型的描述，在 4.8节：

为了便于与非通用遗留代码进行交互，可以使用擦除类型（§4.6）的参数化类型（§4.5）或删除数组类型（§10.1）的元素类型是参数化类型。这种类型被称为原始类型。

GenericRaw< T> 的specs / jls / se7 / html / jls-4.html＃jls-4.6rel =nofollow>擦除为：

  public class GenericRaw {
 public List get（）{...} 
} 
  $ b 
 
 
类型擦除还将构造函数或方法的签名（第8.4.2节）映射到没有参数化类型或类型变量的签名。删除构造函数或方法签名s是一个签名，它包含与s相同的名称和s中给出的所有形式参数类型的删除。 
 
 
类型参数（§8.4.4）和方法的返回类型（§8.4.5），如果构造函数或方法的签名被擦除，它也会被擦除。
 
 
 
I have the following Generic class:
import java.util.ArrayList;
import java.util.List;

public class GenericRaw<T> {
    public List<String> get() {
         return new ArrayList<>();
    }
}
Let us consider the case of its usage:
public class Usage {
    public void doSomething() {
        GenericRaw base = new GenericRaw();
        for (String x : base.get()) {   }
    }
}
For this code Idea doesn't give any compilation error but Java compiler itself does:

  
java: incompatible types 
  required: java.lang.String 
  found: java.lang.Object
Reproducible on JDK 1.6.0_33 as well as on JDK 1.7.0_17.

Can someone help me with the explanation of this issue?

Results of my investigations.
The following variants can be successfully compiled:

public void doSomething() {
    GenericRaw<?> base = new GenericRaw();
    for (String x : base.get()) {   }
}

or even:

public void doSomething() {
    GenericRaw base = new GenericRaw();
    List<String> list = base.get();
    for (String x : list) {   }
}


解决方案

  
Can someone help me with the explanation of this issue?
Sure. It's following the JLS's description of raw types, in section 4.8:

  
To facilitate interfacing with non-generic legacy code, it is possible to use as a type the erasure (§4.6) of a parameterized type (§4.5) or the erasure of an array type (§10.1) whose element type is a parameterized type. Such a type is called a raw type.
The erasure of GenericRaw<T> is:
public class GenericRaw {
    public List get() { ... }
}
because:

  
Type erasure also maps the signature (§8.4.2) of a constructor or method to a signature that has no parameterized types or type variables. The erasure of a constructor or method signature s is a signature consisting of the same name as s and the erasures of all the formal parameter types given in s.
  
  
The type parameters of a constructor or method (§8.4.4), and the return type (§8.4.5) of a method, also undergo erasure if the constructor or method's signature is erased.


                        
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