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只用BIT操作,求解整型和浮点数的绝对值

2013年06月16日 ⁄ 综合 ⁄ 共 818字 ⁄ 字号 评论关闭

Q: Given one integer number, how to solve its absolute with bitwise op, and how about floating number?


A:

Q1: For integer numbers:

1. For positive number, its abs value is still itself
2. For negative number, its abs value will ~(x-1).

How to combine above expression, without any branching, including ternary op(:?).

Code as below:

    #define BITS_PER_BYTE 8
    int myabs(int x)
    {
        int mask = (x >> (BITS_PER_BYTE * sizeof(int) - 1));  // shift right with signed bit
        return (x + mask) ^ mask;
    }

For negative number, mask is -1, while 0 for positive. ^mask op is like ~ bit op.

Q2: For floating number, here take example of float type:

So abs floating number, only needs abs the MSB bit.

    float myfabs(float f)
    {
        int* temp = (int*)&f;
        int mask = ~(1 << (BITS_PER_BYTE * sizeof(int) - 1));  // only MSB bit is 1, other bits are 0
        int out = *temp & mask;  // set MSB bit to 0
        return *((float*)&out);
    }

For double type, representation is similar, totally 64 bits, 11 bits for magnitude(exp), 52 bits for fraction.



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