快速回答
您应该使用equals
方法,因为这是为了执行您想要的比较而实现的。toString()
本身使用迭代器就像equals
但这是一种效率较低的方法。此外,正如@Teepeemm 指出的那样,toString
受元素顺序(基本上是迭代器返回顺序)的影响,因此不能保证为 2 个不同的映射提供相同的输出(特别是如果我们比较两个不同的映射)。
注意/警告:你的问题和我的答案假设实现地图接口的类尊重预期toString
and equals
行为。默认的 java 类会这样做,但需要检查自定义映射类以验证预期的行为。
See: http://docs.oracle.com/javase/7/docs/api/java/util/Map.html http://docs.oracle.com/javase/7/docs/api/java/util/Map.html
boolean equals(Object o)
比较指定对象与此映射是否相等。返回真
如果给定的对象也是一个地图并且两个地图代表同一个
映射。更正式地说,两个映射 m1 和 m2 表示相同
映射 if m1.entrySet().equals(m2.entrySet())。这确保了
equals 方法可以在不同的实现中正常工作
地图界面。
Java 源代码中的实现 (java.util.AbstractMap)
此外,java 本身负责遍历所有元素并进行比较,因此您无需这样做。看看执行情况AbstractMap
它被诸如这样的类使用HashMap
:
// Comparison and hashing
/**
* Compares the specified object with this map for equality. Returns
* <tt>true</tt> if the given object is also a map and the two maps
* represent the same mappings. More formally, two maps <tt>m1</tt> and
* <tt>m2</tt> represent the same mappings if
* <tt>m1.entrySet().equals(m2.entrySet())</tt>. This ensures that the
* <tt>equals</tt> method works properly across different implementations
* of the <tt>Map</tt> interface.
*
* <p>This implementation first checks if the specified object is this map;
* if so it returns <tt>true</tt>. Then, it checks if the specified
* object is a map whose size is identical to the size of this map; if
* not, it returns <tt>false</tt>. If so, it iterates over this map's
* <tt>entrySet</tt> collection, and checks that the specified map
* contains each mapping that this map contains. If the specified map
* fails to contain such a mapping, <tt>false</tt> is returned. If the
* iteration completes, <tt>true</tt> is returned.
*
* @param o object to be compared for equality with this map
* @return <tt>true</tt> if the specified object is equal to this map
*/
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof Map))
return false;
Map<K,V> m = (Map<K,V>) o;
if (m.size() != size())
return false;
try {
Iterator<Entry<K,V>> i = entrySet().iterator();
while (i.hasNext()) {
Entry<K,V> e = i.next();
K key = e.getKey();
V value = e.getValue();
if (value == null) {
if (!(m.get(key)==null && m.containsKey(key)))
return false;
} else {
if (!value.equals(m.get(key)))
return false;
}
}
} catch (ClassCastException unused) {
return false;
} catch (NullPointerException unused) {
return false;
}
return true;
}
比较两种不同类型的地图
toString
比较时惨败TreeMap
and HashMap
though equals
确实正确比较内容。
Code:
public static void main(String args[]) {
HashMap<String, Object> map = new HashMap<String, Object>();
map.put("2", "whatever2");
map.put("1", "whatever1");
TreeMap<String, Object> map2 = new TreeMap<String, Object>();
map2.put("2", "whatever2");
map2.put("1", "whatever1");
System.out.println("Are maps equal (using equals):" + map.equals(map2));
System.out.println("Are maps equal (using toString().equals()):"
+ map.toString().equals(map2.toString()));
System.out.println("Map1:"+map.toString());
System.out.println("Map2:"+map2.toString());
}
Output:
Are maps equal (using equals):true
Are maps equal (using toString().equals()):false
Map1:{2=whatever2, 1=whatever1}
Map2:{1=whatever1, 2=whatever2}