Java：If vs. Switch

Question

我有一段代码用a）我纯粹是为了易读而用b）代替了…

一个）

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;
}

…交换机版本会在所有排列中级联还是跳到一个案例？

编辑：

下面的一些答案是上述方法的替代方法。
我包括以下内容以提供其使用的上下文。

我问上述问题的原因是，凭经验提高了添加单词的速度。

这绝对不是生产代码，并且作为PoC很快就被黑了。

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单词更大的单词集。
失败的原因是我没有考虑仍然需要内存的空引用。 （do！）

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 );
    }
}

Answer 1

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，另一个则稀疏。请参阅JVM规范中有关编译开关的描述。对于枚举，找到序数，然后根据int情况继续执行代码。我不完全确定将如何实施JDK7中switch关于String小功能的建议。

但是，通常在任何明智的JVM中编译频繁使用的代码。优化器并不完全是愚蠢的。不必担心，并遵循通常的启发式方法进行优化。