问一个JAVA代码性能问题

Question

1: HashMap<String, String> test = new HashMap<>();
2: Map<String, String> test = new HashMap<>();

只进行put、get操作
请问1的性能会优于2吗？为什么？

Answer 1

    HashMap<String, String> map1 = new HashMap<>();
    Map<String, String> map2 = new HashMap<>();
    map1.put("a", "b");
    map2.put("a", "b");
    

=>

16  aload_1 [map1]
17  ldc <String "a"> [160]
19  ldc <String "b"> [162]
21  invokevirtual java.util.HashMap.put(java.lang.Object, java.lang.Object) : java.lang.Object [164]
24  pop
25  aload_2 [map2]
26  ldc <String "a"> [160]
28  ldc <String "b"> [162]
30  invokeinterface java.util.Map.put(java.lang.Object, java.lang.Object) : java.lang.Object [168] [nargs: 3]

In the test here, http://bobah.net/book/export/html/55
invokeinterface may be 38% slower

http://stackoverflow.com/questions/1504633/what-is-the-point-of-invokeinterface
Explanation here

Answer 2

For this question, the general answer is "about the same, there is no difference";
The more extreme answer is "the performance of 2 is slightly better than 1";

Code below:

HashMap<String, String> m1 = new HashMap<>();
        m1.put("test", "test");
        m1.get("test");

        Map<String, String> m2 = new HashMap<>();
        m2.put("test", "test");
        m2.get("test");

The corresponding instructions after being compiled into bytecode are:

0: new           #16                 // class java/util/HashMap
         3: dup           
         4: invokespecial #18                 // Method java/util/HashMap."<init>":()V
         7: astore_1      
         8: aload_1       
         9: ldc           #19                 // String test
        11: ldc           #19                 // String test
        13: invokevirtual #21                 // Method java/util/HashMap.put:(Ljava/lang/Object;Ljava/lang/Object;)Ljava/lang/Object;
        16: pop           
        17: aload_1       
        18: ldc           #19                 // String test
        20: invokevirtual #25                 // Method java/util/HashMap.get:(Ljava/lang/Object;)Ljava/lang/Object;
        23: pop           
        24: new           #16                 // class java/util/HashMap
        27: dup           
        28: invokespecial #18                 // Method java/util/HashMap."<init>":()V
        31: astore_2      
        32: aload_2       
        33: ldc           #19                 // String test
        35: ldc           #19                 // String test
        37: invokeinterface #29,  3           // InterfaceMethod java/util/Map.put:(Ljava/lang/Object;Ljava/lang/Object;)Ljava/lang/Object;
        42: pop           
        43: aload_2       
        44: ldc           #19                 // String test
        46: invokeinterface #32,  2           // InterfaceMethod java/util/Map.get:(Ljava/lang/Object;)Ljava/lang/Object;
        51: pop           
        52: return

It can be seen that the put/get operation of the map in case 1 is completed with the invokevirtual instruction;
And the put/get operation of the map in case 2 is completed with the invokeinterface instruction;

In terms of implementation, the performance of invokevirtual is slightly better than invokeinterface, so if we have to say who has better performance, it would be 2;

Finally, as a reminder, in the process of java programming, any jvm instruction should be regarded as having almost the same constant-level time overhead, even if it is invokedynamic. Only in this way can we bring unity to our upper-level algorithms and logic optimization , non-interference perspective;
It is irrational to change the way Java code is written in order to pick out jvm instructions in a sharp way, and its conclusion is also unstable - it may change with the upgrading of jvm, and for this purpose The code changes brought about by this kind of "performance improvement" lead to reduced readability and maintainability, which is not worth the gain

Answer 3

Based on operational methodology, 2 performs better than 1