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Java通用问题(Java Generic Question)

下面的代码编译,但如果我取消注释注释行,它不会,我很困惑为什么。 HashMap确实扩展了AbstractMap,并且声明map的第一行编译正常。

import java.util.AbstractMap;
import java.util.HashMap;
import java.util.Map;

public class Test {

    public static void main(String args[]) {
        Map<String, ? extends AbstractMap<String, String>> map = new HashMap<String, HashMap<String, String>>();
        //map.put("one", new HashMap<String, String>());
    }
}

而且,我知道“正确的方式”是这样的:

import java.util.HashMap;
import java.util.Map;

public class Test {

    public static void main(String args[]) {
        Map<String, Map<String, String>> map = new HashMap<String, Map<String, String>>();
        map.put("one", new HashMap<String, String>());
    }
}

The following code compiles but if I uncomment the commented line, it does not and I am confused why. HashMap does extend AbstractMap and the first line where map is declared compiles fine.

import java.util.AbstractMap;
import java.util.HashMap;
import java.util.Map;

public class Test {

    public static void main(String args[]) {
        Map<String, ? extends AbstractMap<String, String>> map = new HashMap<String, HashMap<String, String>>();
        //map.put("one", new HashMap<String, String>());
    }
}

And, I know the "right way" is this:

import java.util.HashMap;
import java.util.Map;

public class Test {

    public static void main(String args[]) {
        Map<String, Map<String, String>> map = new HashMap<String, Map<String, String>>();
        map.put("one", new HashMap<String, String>());
    }
}

原文:https://stackoverflow.com/questions/5974065
更新时间:2023-04-30 11:04

最满意答案

你是对的。 检测无限递归而不限制模板元编程上的递归堆栈帧意味着找到停止问题的替代解决方案。

理论上有几种特殊情况是可检测的。 例如,如果您可以确保递归的引用透明性,并且最后一个函数调用接收到与实际调用相同的参数,那么您将进行无限递归调用。 C ++不提供模板元编程的参考透明度保证。


Your are correct. Detecting an infinite recursion without limiting the recursion stack frames on template meta programming would mean finding an alternative solution to the halting problem.

There are a few special cases which are, in theory, detectable. For example, if you can ensure referential transparency on the recursion and if the last function call receives the same parameters as the actual one, you are on an infinite recursive call. C++ offers no referential transparency warranty on template meta programming.

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