[Happy Coding] 只用BIT操作，求解整型和浮点数的绝对值

最新推荐文章于 2021-08-09 13:01:11 发布

原创最新推荐文章于 2021-08-09 13:01:11 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

C++语言专栏收录该内容

21 篇文章

订阅专栏

本文介绍了如何使用位运算来计算整数和浮点数的绝对值，针对整数提供了特定的位运算方法，并解释了如何在浮点数上应用类似的技巧。详细步骤包括对整数进行位移操作以获取符号位，并结合位运算符实现非分支操作；对于浮点数，则聚焦于获取最高有效位并调整其值。

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

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.