C 到 Swift 的按位运算

Question

我在 C 语言中有这些函数（来自 Cactus Kev 的扑克评估器）：

unsigned find_fast(unsigned u)
{
    unsigned a, b, r;
    u += 0xe91aaa35;
    u ^= u >> 16;
    u += u << 8;
    u ^= u >> 4;
    b  = (u >> 8) & 0x1ff;
    a  = (u + (u << 2)) >> 19;
    r  = a ^ hash_adjust[b];
    return r;
}

int eval_5hand_fast(int c1, int c2, int c3, int c4, int c5)
{
    int q = (c1 | c2 | c3 | c4 | c5) >> 16;
    short s;
    if (c1 & c2 & c3 & c4 & c5 & 0xf000)
        return flushes[q]; 
    if ((s = unique5[q]))
        return s;          
    return hash_values[find_fast((c1 & 0xff) * (c2 & 0xff) * (c3 & 0xff) * (c4 & 0xff) * (c5 & 0xff))];
}

并希望将它们转换为 Swift：

func eval_5hand_fast(c1: Int, c2: Int, c3: Int, c4: Int, c5: Int) -> Int {

    var q: Int = (c1 | c2 | c3 | c4 | c5) >> 16
    var s: Int8
    if c1 & c2 & c3 & c4 & c5 & 0xf000 {
        return flushes[q]
    }
    if (s = unique5[q]) {
        return s
    }
    return hash_values[find_fast((c1 & 0xff) * (c2 & 0xff) * (c3 & 0xff) * (c4 & 0xff) * (c5 & 0xff))]
}

func find_fast(u: UInt) -> UInt {
    var a, b, r: UInt
    u += 0xe91aaa35
    u ^= u >> 16
    u += u << 8
    u ^= u >> 4
    b  = (u >> 8) & 0x1ff
    a  = (u + (u << 2)) >> 19
    r  = a ^ hash_adjust[b]
    return r;
}

我只是在学习 Swift，并没有真正做太多的按位运算，所以请多多包涵。我试图自己解决这些问题，但无济于事。我在使用 Swift 时遇到的错误包括：

1. if c1 & c2 & c3 & c4 & c5 & 0xf000
   // ERROR: Type 'Int' does not conform to protocol 'BooleanType'

2. if (s = unique5[q])
   // ERROR: Type '()' does not conform to protocol 'BooleanType'

3. return hash_values[find_fast((c1 & 0xff) * (c2 & 0xff) * (c3 & 0xff) * (c4 & 0xff) * (c5 & 0xff))]
   // ERROR: Cannot find an overload for '*' that accepts the supplied arguments

4. u += 0xe91aaa35
   // ERROR: Cannot invoke '+=' with an argument list of type '(UInt, IntegerLiteralConvertible)'

5. u ^= u >> 16
   // ERROR: Cannot invoke '>>' with an argument list of type '(UInt, $T5)'

6. u += u << 8
   // ERROR: Cannot invoke '+=' with an argument list of type '(UInt, $T5)'
   

7. u ^= u >> 4
   // ERROR: Cannot invoke '>>' with an argument list of type '(UInt, $T5)'

我收到第一个错误（它不是布尔值），但是我不确定如何解决它，因为我不完全确定 C 版本的要求是什么，因为我不太熟悉使用位和掩码。其他错误我真的不知道该怎么办。

Answer 1

当涉及到整数时，C 中任何非 0 都被视为 true。Swift 需要一个布尔值，因此您必须添加 != 0。例如：

C:     if c1 & c2 & c3 & c4 & c5 & 0xf000
Swift: if c1 & c2 & c3 & c4 & c5 & 0xf000 != 0

C:     if (s = unique5[q])
Swift: if let s = unique5[q] where s != 0

试试这个：

func eval_5hand_fast(c1: Int, c2: Int, c3: Int, c4: Int, c5: Int) -> Int {

    var q: Int = (c1 | c2 | c3 | c4 | c5) >> 16
    var s: Int8
    if c1 & c2 & c3 & c4 & c5 & 0xf000 != 0 {
        return flushes[q]
    }
    if let s = unique5[q] where s != 0 {
        return s
    }
    return hash_values[find_fast((c1 & 0xff) * (c2 & 0xff) * (c3 & 0xff) * (c4 & 0xff) * (c5 & 0xff))]
}

func find_fast(var u: UInt) -> UInt {
    var a, b, r: UInt
    u += 0xe91aaa35
    u ^= u >> 16
    u += u << 8
    u ^= u >> 4
    b  = (u >> 8) & 0x1ff
    a  = (u + (u << 2)) >> 19
    r  = a ^ hash_adjust[b]
    return r;
}