为什么 Java 7 中的 switch 语句比 String 的 if else 更快? [英] Why is the switch statement faster than if else for String in Java 7?

问题描述

在 Java 7 中，string 对象可以在 switch 语句的表达式中.有人可以从 官方文档?

In Java 7 a string object can be in the expression of a switch statement. Can someone explain the below statement from official documentation?

Java 编译器从使用 String 对象的 switch 语句生成的字节码通常比从链式 if-then-else 语句生成的字节码效率更高.

The Java compiler generates generally more efficient bytecode from switch statements that use String objects than from chained if-then-else statements.

推荐答案

Java代码

有两个版本的类，例如

Java Code

Having two versions of a class, e.g.

使用 if-then-else:

public class IfThenElseClass {
    public static void main(String[] args) {
        String str = "C";
        if ("A".equals(str)) {

        } else if ("B".equals(str)) {

        } else if ("C".equals(str)) {

        }
    }
}

带开关:

public class SwitchClass {
    public static void main(String[] args) {
        String str = "C";
        switch (str) {
            case "A":
                break;
            case "B":
                break;
            case "C":
                break;
        }
    }
}

字节码

让我们看一下字节码.获取 if-then-else 版本的字节码:

Compiled from "CompileSwitch.java"
public class CompileSwitch {
  public CompileSwitch();
    Code:
       0: aload_0
       1: invokespecial #8  // Method java/lang/Object."<init>":()V
       4: return

  public static void main(java.lang.String[]);
    Code:
       0: ldc           #16 // String C
       2: astore_1
       3: ldc           #18 // String A
       5: aload_1
       6: invokevirtual #20 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
       9: ifne          28
      12: ldc           #26 // String B
      14: aload_1
      15: invokevirtual #20 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
      18: ifne          28
      21: ldc           #16 // String C
      23: aload_1
      24: invokevirtual #20 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
      27: pop
      28: return
}

获取switch版本的字节码:

Compiled from "CompileSwitch.java"
public class CompileSwitch {
  public CompileSwitch();
    Code:
       0: aload_0
       1: invokespecial #8 // Method java/lang/Object."<init>":()V
       4: return

  public static void main(java.lang.String[]);
    Code:
       0: ldc           #16 // String C
       2: astore_1
       3: aload_1
       4: dup
       5: astore_2
       6: invokevirtual #18 // Method java/lang/String.hashCode:()I
       9: lookupswitch  { // 3
                    65: 44
                    66: 56
                    67: 68
               default: 77
          }
      44: aload_2
      45: ldc           #24 // String A
      47: invokevirtual #26 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
      50: ifne          77
      53: goto          77
      56: aload_2
      57: ldc           #30 // String B
      59: invokevirtual #26 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
      62: ifne          77
      65: goto          77
      68: aload_2
      69: ldc           #16 // String C
      71: invokevirtual #26 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
      74: ifne          77
      77: return
}

结论

• 在第一个版本中，通过对每个条件调用 equals 方法来比较字符串，直到找到为止.

  Conclusion

  • In the first version compares the string by calling the equals method for each condition, until it is found.

    在第二个版本中是获取第一个hashCode的字符串.然后将其与每个case的值hashCode进行比较.请参阅 lookupswitch.如果这些值中的任何一个重复，则恰好运行 case 的代码.否则，调用绑定案例的 equals 方法.这比仅调用 equals 方法要快得多.

    In the second version is obtained first hashCode of the string. Then this is compared with the values ​​hashCode each case. See the lookupswitch. If any of these values ​​is repeated just happens to run the code for the case. Otherwise, call the equals method of the cases tied. This is much faster than ever call the equals method only.

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