Question

我有一段代码与a）我用b）代替纯粹是为了易读......

A）

if ( WORD[ INDEX ] == 'A' ) branch = BRANCH.A;
/* B through to Y */
if ( WORD[ INDEX ] == 'Z' ) branch = BRANCH.Z;

b）中

switch ( WORD[ INDEX ] ) {
    case 'A' : branch = BRANCH.A; break;
    /* B through to Y */
    case 'Z' : branch = BRANCH.Z; break;
}

...交换机版本是否会级联所有排列或跳转到案例？

编辑：

下面的一些答案涉及上述方法的替代方法。
我已提供以下内容以提供其使用的背景信息。

我问的问题，上面的问题是因为增加词语的速度在经验上有所改善。

这不是生产代码，而是作为PoC快速入侵。

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词语更大的词汇。
失败的原因是我没有考虑到仍需要内存的空引用。（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 );
    }
}

Answer 1

不要担心性能;使用最能表达您正在做的事情的语法。只有在你（a）表现出性能不足之后; （b）将其本地化为相关例程，只有这样你才应该担心表现。对于我的钱，案例语法在这里更合适。

Answer 2

在字节码中有两种形式的切换：tableswitch和lookupswitch。一个假设密集的密钥，另一个密集。请参阅compiling switch in the JVM spec的说明。对于枚举，找到序号，然后代码继续作为int案例。我不完全确定如何实现JDK7中提议的switch String小功能。

但是，大量使用的代码通常在任何合理的JVM中编译。优化者并非完全愚蠢。不要担心，并按照通常的启发式方法进行优化。

Answer 3

看起来你已经枚举了这些值，所以也许枚举是按顺序的？

enum BRANCH {
  A,B, ... Y,Z;
}

然后在你的代码中：

BRANCH branch = BRANCH.valueOf( WORD[ INDEX ] );

此外，根据“A”的对象标识，您的代码中可能存在"A" == "A"可能为false的错误。

Answer 4

对你的问题不是一个完全的答案，但你的代码中确实存在一个错误，你应该在每个案例后休息一下：

switch ( WORD[ INDEX ] ) {
    case 'A' : branch = BRANCH.A; break;
    /* B through to Y */
    case 'Z' : branch = BRANCH.Z; break;
}

我不认为这里的性能差异太大，但如果你真的关心性能，如果这个代码经常执行，那么这是另一种选择：

// Warning, untested code.
BRANCH[] branchLookUp = {BRANCH.A, BRANCH.B, ..., BRANCH.Z};

branch = branchLookUp[WORD[INDEX] - 'A'];

请务必将其封装并记录下来。

Answer 5

老实说，在这种情况下，我认为表现并不重要。这完全取决于编译器及其优化。

Answer 6

如果你有一个带有连续积分值的switch语句，这取决于语言，它可以优化到分支表，这非常快。反正它并不慢！

此外，使用if / else if对于诸如此类案例互相排斥的情况是否会有所改进。在匹配A之后再做25次检查是没有意义的。

但基本上，任何性能差异都可以忽略不计，你应该使用最正确的语法，在本例中是switch语句。确保将您的案件与休息分开。

Answer 7

这是一种避免所有案例陈述的方法。

import java.util.HashMap;

public class Dictionary {
    private static Dictionary                       ROOT;
    private boolean                                 terminus;
    private final HashMap<Character, Dictionary>    dictionaries    = new HashMap<Character, Dictionary>();

    private void ensureBranch(char c) {
        if (getBranch(c) != null)
            return;
        dictionaries.put(c, new Dictionary());
    }

    private Dictionary getBranch(char c) {
        return dictionaries.get(c);
    }

    public static boolean is(final String string) {
        ensureRoot();
        return is(chars(string), ROOT, 0, string.length() - 1);
    }

    public static void add(final String... strings) {
        ensureRoot();
        for (final String string : strings)
            add(chars(string), ROOT, 0, string.length() - 1);
    }

    private static void ensureRoot() {
        if (ROOT == null)
            ROOT = new Dictionary();
    }

    private static char[] chars(final String string) {
        return string.toUpperCase().toCharArray();
    }

    private Dictionary() {
        this.terminus = false;
    }

    private static void add(final char[] word, final Dictionary dictionary, final int index, final int limit) {
        Dictionary branch = getBranch(word, dictionary, index);
        if (index == limit)
            branch.terminus = true;
        else
            add(word, branch, index + 1, limit);
    }

    private static Dictionary getBranch(final char[] word, final Dictionary dictionary, final int index) {
        final char c = word[index];
        dictionary.ensureBranch(c);
        return dictionary.getBranch(c);
    }

    private static boolean is(final char[] word, final Dictionary dictionary, final int index, final int limit) {
        Dictionary branch = dictionary.getBranch(word[index]);
        if (branch == null)
            return false;
        if (index == limit)
            return branch.terminus;
        return is(word, branch, index + 1, limit);
    }
}

Answer 8

我知道这不是你要求的，但你不是这样做的吗？

public class Dictionary {
    private static final Set<String> WORDS = new HashSet<String>();

    public static void add(final String... STRINGS) {
        for (String str : STRINGS) {
            WORDS.add(str.toUpperCase());
        }
    }

    public static boolean is(final String STRING) {
        return WORDS.contains(STRING.toUpperCase());
    }
}

您只是在寻找更高效的内存吗？

Answer 9

switch语句应使用哈希来选择要转到的大小写。从那里开始，如果没有break语句，也会运行每个后续案例。例如，使用您的代码，如果您打开X，它将立即转到X，然后是Y，然后转到Z.请参阅Java Tutorial。

Answer 10

假设编译器找不到任何聪明的东西，switch应该是对数的并且if是线性的。但是长switch es很难阅读并且容易出错 - 如上所述，你上面的开关没有任何中断，并且它会在所有情况下失败。

为什么不预先填充Map，而只使用Map.get()？

private static final Map<Char, Whatever> BRANCHES = Collections.unmodifiableMap(new HashMap<Char, Whatever>() {{
    put('A', BRANCH.A);
    ...
    put('Z', BRANCH.Z);
}}

public void getBranch(char[] WORD, int INDEX) {
    return BRANCHES.get(WORD[INDEX]);
}

如上所述，如果BRANCH是Enum，则此行为应该恰好位于Enum。

（这里有什么WORD，INDEX和BRANCH？从名称来看，它们应该是常量，但你不能真正拥有常量数组 - 内容是alwyas modifiable;创建一个常量“struct”没有多大意义;并且基于常量的iffing或切换肯定没有多大意义....）

Answer 11

我认为这更像是一个关于风格而不是表现的问题。我认为在这种情况下，switch语句比if更合适。我不确定性能有多大差异。

Java：If vs. Switch

...交换机版本是否会级联所有排列或跳转到案例？

11 个答案: