Java:如果与 Switch [英] Java: If vs. Switch
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问题描述
我有一段带有 a) 的代码，我将其替换为 b) 纯粹是为了便于阅读......
a)
if ( WORD[ INDEX ] == 'A' ) branch = BRANCH.A;
/* B through to Y */
if ( WORD[ INDEX ] == 'Z' ) branch = BRANCH.Z;
b)
switch ( WORD[ INDEX ] ) {
case 'A' : branch = BRANCH.A; break;
/* B through to Y */
case 'Z' : branch = BRANCH.Z; break;
}
下面的一些答案是关于上述方法的替代方法.
我已经包含以下内容以提供其使用的上下文.
我之所以问上面的问题，是因为经验性地提高了添加单词的速度.
这绝不是生产代码，并且作为 PoC 被迅速破解.
以下似乎是对思想实验失败的确认.
我可能需要比我目前使用的更大的词库.
失败是因为我没有考虑仍然需要内存的空引用.(doh！)
public class Dictionary {
private static Dictionary ROOT;
private boolean terminus;
private Dictionary A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z;
private static Dictionary instantiate( final Dictionary DICTIONARY ) {
return ( DICTIONARY == null ) ? new Dictionary() : DICTIONARY;
}
private Dictionary() {
this.terminus = false;
this.A = this.B = this.C = this.D = this.E = this.F = this.G = this.H = this.I = this.J = this.K = this.L = this.M = this.N = this.O = this.P = this.Q = this.R = this.S = this.T = this.U = this.V = this.W = this.X = this.Y = this.Z = null;
}
public static void add( final String...STRINGS ) {
Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT );
for ( final String STRING : STRINGS ) Dictionary.add( STRING.toUpperCase().toCharArray(), Dictionary.ROOT , 0, STRING.length() - 1 );
}
private static void add( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) {
Dictionary branch = null;
switch ( WORD[ INDEX ] ) {
case 'A' : branch = BRANCH.A = Dictionary.instantiate( BRANCH.A ); break;
case 'B' : branch = BRANCH.B = Dictionary.instantiate( BRANCH.B ); break;
case 'C' : branch = BRANCH.C = Dictionary.instantiate( BRANCH.C ); break;
case 'D' : branch = BRANCH.D = Dictionary.instantiate( BRANCH.D ); break;
case 'E' : branch = BRANCH.E = Dictionary.instantiate( BRANCH.E ); break;
case 'F' : branch = BRANCH.F = Dictionary.instantiate( BRANCH.F ); break;
case 'G' : branch = BRANCH.G = Dictionary.instantiate( BRANCH.G ); break;
case 'H' : branch = BRANCH.H = Dictionary.instantiate( BRANCH.H ); break;
case 'I' : branch = BRANCH.I = Dictionary.instantiate( BRANCH.I ); break;
case 'J' : branch = BRANCH.J = Dictionary.instantiate( BRANCH.J ); break;
case 'K' : branch = BRANCH.K = Dictionary.instantiate( BRANCH.K ); break;
case 'L' : branch = BRANCH.L = Dictionary.instantiate( BRANCH.L ); break;
case 'M' : branch = BRANCH.M = Dictionary.instantiate( BRANCH.M ); break;
case 'N' : branch = BRANCH.N = Dictionary.instantiate( BRANCH.N ); break;
case 'O' : branch = BRANCH.O = Dictionary.instantiate( BRANCH.O ); break;
case 'P' : branch = BRANCH.P = Dictionary.instantiate( BRANCH.P ); break;
case 'Q' : branch = BRANCH.Q = Dictionary.instantiate( BRANCH.Q ); break;
case 'R' : branch = BRANCH.R = Dictionary.instantiate( BRANCH.R ); break;
case 'S' : branch = BRANCH.S = Dictionary.instantiate( BRANCH.S ); break;
case 'T' : branch = BRANCH.T = Dictionary.instantiate( BRANCH.T ); break;
case 'U' : branch = BRANCH.U = Dictionary.instantiate( BRANCH.U ); break;
case 'V' : branch = BRANCH.V = Dictionary.instantiate( BRANCH.V ); break;
case 'W' : branch = BRANCH.W = Dictionary.instantiate( BRANCH.W ); break;
case 'X' : branch = BRANCH.X = Dictionary.instantiate( BRANCH.X ); break;
case 'Y' : branch = BRANCH.Y = Dictionary.instantiate( BRANCH.Y ); break;
case 'Z' : branch = BRANCH.Z = Dictionary.instantiate( BRANCH.Z ); break;
}
if ( INDEX == INDEX_LIMIT ) branch.terminus = true;
else Dictionary.add( WORD, branch, INDEX + 1, INDEX_LIMIT );
}
public static boolean is( final String STRING ) {
Dictionary.ROOT = Dictionary.instantiate( Dictionary.ROOT );
return Dictionary.is( STRING.toUpperCase().toCharArray(), Dictionary.ROOT, 0, STRING.length() - 1 );
}
private static boolean is( final char[] WORD, final Dictionary BRANCH, final int INDEX, final int INDEX_LIMIT ) {
Dictionary branch = null;
switch ( WORD[ INDEX ] ) {
case 'A' : branch = BRANCH.A; break;
case 'B' : branch = BRANCH.B; break;
case 'C' : branch = BRANCH.C; break;
case 'D' : branch = BRANCH.D; break;
case 'E' : branch = BRANCH.E; break;
case 'F' : branch = BRANCH.F; break;
case 'G' : branch = BRANCH.G; break;
case 'H' : branch = BRANCH.H; break;
case 'I' : branch = BRANCH.I; break;
case 'J' : branch = BRANCH.J; break;
case 'K' : branch = BRANCH.K; break;
case 'L' : branch = BRANCH.L; break;
case 'M' : branch = BRANCH.M; break;
case 'N' : branch = BRANCH.N; break;
case 'O' : branch = BRANCH.O; break;
case 'P' : branch = BRANCH.P; break;
case 'Q' : branch = BRANCH.Q; break;
case 'R' : branch = BRANCH.R; break;
case 'S' : branch = BRANCH.S; break;
case 'T' : branch = BRANCH.T; break;
case 'U' : branch = BRANCH.U; break;
case 'V' : branch = BRANCH.V; break;
case 'W' : branch = BRANCH.W; break;
case 'X' : branch = BRANCH.X; break;
case 'Y' : branch = BRANCH.Y; break;
case 'Z' : branch = BRANCH.Z; break;
}
if ( branch == null ) return false;
if ( INDEX == INDEX_LIMIT ) return branch.terminus;
else return Dictionary.is( WORD, branch, INDEX + 1, INDEX_LIMIT );
}
}
推荐答案
在字节码中有两种形式的switch:tableswitch和lookupswitch.一个假设一组密集的键，另一个是稀疏的.参见JVM规范中编译开关的描述.对于枚举，找到序数，然后代码继续 int 的情况.我不完全确定 JDK7 中 String 小特性上提出的 switch 将如何实现.
In bytecode there are two forms of switch: tableswitch and lookupswitch. One assumes a dense set of keys, the other sparse. See the description of compiling switch in the JVM spec. For enums, the ordinal is found and then the code continues as the int case. I am not entirely sure how the proposed switch on String little feature in JDK7 will be implemented.
但是，大量使用的代码通常在任何合理的 JVM 中编译.优化器并不完全愚蠢.不用担心，按照通常的启发式进行优化.
However, heavily used code is typically compiled in any sensible JVM. The optimiser is not entirely stupid. Don't worry about it, and follow the usual heuristics for optimisation.
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