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Java generics problem with Map

开发者 https://www.devze.com 2023-04-06 20:30 出处:网络
I have thi开发者_高级运维s which should not compile. public boolean foo(final Map<String, String> map) {

I have thi开发者_高级运维s which should not compile.

public boolean foo(final Map<String, String> map) {
     map.get(1);   // No error here?
}


The get method takes an Object, so the 1 gets autoboxed to an Integer and it compiles just fine. You'll always get null back as a response though.

This is done for backwards compatability.

// Ideal Map:
public V get(K key) {
 // impl here
}

// what we have
public V get(Object key) {
 // impl here
}

Example for scorpian

import java.util.*;
class ScorpMain {

    /* This interface just has a single method on it returning an int */
    static interface SomeInterface {
        int foo();
    }

    /**
     * ExampleKey has an int and a string. It considers itself to be equal to
     * another object if that object implements SomeInterface and the two have
     * equal foo values. It believes this is sufficient, as its sole purpose is
     * to calculate this foo value.
     */
    static class ExampleKey implements SomeInterface {
        int i;
        String s;
        ExampleKey(int i, String s) { this.i = i; this.s = s; }
        public int foo() { return i * s.length(); }
        public int hashCode() { return i ^ s.hashCode(); }
        // notice - equals takes Object, not ExampleKey
        public boolean equals(Object o) {
            if(o instanceof SomeInterface) {
                SomeInterface so = (SomeInterface)o;
                System.out.println(so.foo() + " " + foo());
                return so.foo() == foo();
            }
            return false;
        }
    }

    /**
     * The ImposterKey stores it's foo and hash value. It has no calculation 
     * involved. Note that its only relation to ExampleKey is that they happen
     * to have SomeInterface.
     */
    static class ImposterKey implements SomeInterface {
        int foo;
        int hash;
        ImposterKey(int foo, int hash) { this.foo = foo; this.hash = hash; }
        public int foo() { return foo; }
        public boolean equals(Object o) {
                SomeInterface so = (SomeInterface)o;
                return so.foo() == foo();
        }
        public int hashCode() { return hash; }
    }

    /**
     * In our main method, we make a map. We put a key into the map. We get the
     * data from the map to prove we can get it out. Then we make an imposter, 
     * and get the data based on that. 
     * The moral of the story is, Map.get isn't sacred: if you can trick it 
     * into thinking that the object inside is equal to the object you give it 
     * in both equality and hash, it will give you the resulting object. It 
     * doesn't have anything to do with the type except that a given type is 
     * unlikely to be equal to another object that isn't of the given type.
     * This may seem like a contrived example, and it is, but it is something 
     * to be conscious of when dealing with maps. It's entirely possible you 
     * could get the same data and not realize what you're trying to do. Note 
     * you cannot ADD the imposter, because that IS checked at compile time.
     */
    public static void main(String[] args) {
        Map<ExampleKey,String> map = new HashMap<ExampleKey,String>();
        ExampleKey key = new ExampleKey(1337,"Hi");
        map.put(key,"Awesome!");
        String get = map.get(key);
        System.out.println(get); // we got awesome
        ImposterKey imposter = new ImposterKey(2674,3096); // make an imposter
        String fake = map.get(imposter);
        System.out.println(fake); // we got awesome again!
    }
}


As of Java 6 the line you have written is exactly equivalent to:

public boolean foo(final Map<String, String> map) {
    map.get(Integer.valueOf(1)); // compiles fine of course!
}

Which they call "autoboxing". On the flipside there "autounboxing", ie:

public int bar(final Map<String, Integer> map) {
    return map.get("bar");
}

This line is equivalent to:

return map.get("bar").intValue();

which sometimes leads to a rather mysterious looking NullPointerException when the requested key is not found in the map. This is a common trap for programmers new to this concept.

Hope that helps.


I have this which should not compile.

Why not?

The definition of .get() says that it finds the key such that the given object is equal to the key. Yes, in this case we know that the object is an Integer, and Integer's .equals() method always returns false for a String object. But in general for two objects of different classes we do not know that. In fact, there are cases in the Java library where objects of different classes are required to be equal; for example, java.util.List requires lists to be equal if their contents are equal and in the same order, regardless of class.

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