Question

我必须重新编写一个方法，仅使用按位运算符和没有条件语句来查明x是否小于或等于y。到目前为止我有这个：

int isLessOrEqual(int x, int y)
{
    int a = x + (~y) + 1;
    return ((a&0x80000000) >> 31);
}

但我不知道如何防范溢出？任何人都可以伸出援助之手吗？

Answer 1

[编辑]错误的解决方案，因为它使用条件。我会暂时搁置它，因为它可能会提供见解。

假设我们知道int是4字节2的恭维。

if (x == y) return 1;
int SignMask = 0x80000000;
// if x & y have different signs ...
if ((x & SignMask) != (y & SignMask)) {
  return !!(x & SignMask);  
}
// Continue with your original code knowing overflow will not happen.

Answer 2

仅仅为了练习，下面的扭曲代码将没有任何“+”或条件语句（除非你考虑转换为布尔条件）。

#define MSB 0x80000000
typedef bool (*RecurseFunc)(unsigned int, unsigned int);

bool EndRecursion(unsigned int ux, unsigned int uy)
{
    // stop recursion - ux cannot be smaller than uy
    return false;
}

bool CompareMSB(unsigned int ux, unsigned int uy)
{
    // jump table, may make global...
    RecurseFunc Handler[2] = {EndRecursion, CompareMSB}; 

    // yMsbBigger == MSB only iff Y"MSB==1 and X:MSB==0
    unsigned int yMsbBigger = ((~ux) & uy) & MSB; 
    // differentMsb will be MSB only iff the MSB of Y different MSB of x
    unsigned int differentMsb = (uy ^ ux) & MSB; 

    // prepare to check on the next bit - shift it to MSB position
    ux <<= 1;
    uy <<= 1;

    // we may want to check the remaining bits if ux,uy had same signs and 
    //     the remaining bits of y are not all 0
    bool mayRecurse = ((bool)uy) && (!(bool)differentMsb); 

    // C will not evaluate the second expression if the first is true
    // the second expression will "EndRecursion" if we may not recurse - 
    //     otherwise recurse
    return (((bool)yMsbBigger) || Handler[mayRecurse](ux, uy));
}

bool isYGreater(int x, int y)
{
    // we reverse the sign bits so positive value will have MSB=1, making it 
    //     "greater" then negative value's MSB
    unsigned int xbits = (unsigned int)x ^ MSB;
    unsigned int ybits = (unsigned int)y ^ MSB;

    return (CompareMSB(xbits, ybits));
}

如何仅使用按位运算符来防止C中的位溢出

2 个答案: