Question

我在编译开关时阅读JVM specification，并对如何编译String上的switch语句感兴趣。这是我检查的测试方法（JDK1.7.0_40）：

static int test(String i) {
    switch (i) {
        case "a":  return  -100;
        case "45b":  return  1;
        case "c":  return  2;
        default: return -1;
    }
}

我希望这个方法能够在string的hashCode上编译成简单的lookupswitch，但突然间

static int test(java.lang.String);
Code:
   0: aload_0
   1: astore_1
   2: iconst_m1
   3: istore_2
   4: aload_1
   5: invokevirtual #6         // Method java/lang/String.hashCode:()I
   8: lookupswitch  { // 3
                97: 44
                99: 72
             51713: 58
           default: 83
      }
  44: aload_1
  45: ldc           #7 // String a
  47: invokevirtual #8 // Method java/lang/String.equals:(Ljava/lang/Object;)Z
  50: ifeq          83
  53: iconst_0
  54: istore_2
  55: goto          83
  58: aload_1
  59: ldc           #9  // String 45b
  61: invokevirtual #8  // Method java/lang/String.equals:(Ljava/lang/Object;)Z
  64: ifeq          83
  67: iconst_1
  68: istore_2
  69: goto          83
  72: aload_1
  73: ldc           #10 // String c
  75: invokevirtual #8  // Method java/lang/String.equals:(Ljava/lang/Object;)Z
  78: ifeq          83
  81: iconst_2
  82: istore_2
  83: iload_2
  84: tableswitch   { // 0 to 2
                 0: 112
                 1: 115
                 2: 117
           default: 119
      }
 112: bipush        -100
 114: ireturn
 115: iconst_1
 116: ireturn
 117: iconst_2
 118: ireturn
 119: iconst_m1
 120: ireturn

正如您所看到的，在第一个lookupswitch的分支中，JVM并没有为后续的tableswitch（第84行）做真正的工作生成索引。
Tableswitch应该能够快速工作，因此不会带来很多额外的工作。但无论如何，产生额外开关的目的是什么？的更新
我理解hashCode冲突的可能性。我想说的是，编译器可以将所有实际工作从后续的tableswitch移到first，然后使用ifeq跳转到所有switch分支的末尾，而不是后续的tableswitch。 所以我在这里看到一个可能的答案：在第一个开关编译器尝试根据已知的案例数量为ifeq跳转预先计算标签，但我不确定这是唯一的原因。

的 UPDATE2
正如 @ericbn 建议的那样，我尝试编译

switch (i) { 
   case 97: return -100; 
   case 51713: return 1; 
   case 99: return 2; 
   default: return -1;
}

用i作为int，编译器给了我简单的lookupswitch。

Answer 1

来自javac source code：

的引用

         * The general approach used is to translate a single
         * string switch statement into a series of two chained
         * switch statements: the first a synthesized statement
         * switching on the argument string's hash value and
         * computing a string's position in the list of original
         * case labels, if any, followed by a second switch on the
         * computed integer value.  The second switch has the same
         * code structure as the original string switch statement
         * except that the string case labels are replaced with
         * positional integer constants starting at 0.
         *
         * The first switch statement can be thought of as an
         * inlined map from strings to their position in the case
         * label list.  An alternate implementation would use an
         * actual Map for this purpose, as done for enum switches.
         *
         * With some additional effort, it would be possible to
         * use a single switch statement on the hash code of the
         * argument, but care would need to be taken to preserve
         * the proper control flow in the presence of hash
         * collisions and other complications, such as
         * fallthroughs.  Switch statements with one or two
         * alternatives could also be specially translated into
         * if-then statements to omit the computation of the hash
         * code.

为什么要将String编译切换为两个开关

1 个答案: