快速检查两个 int32_t 之间的差异是 1 还是 0

Question

一个直接的方法是

uint32_t diff = abs(num1 - num2);
bool isZeroOrOne = (diff == 0 || diff == 1);

或者简单地检查所有可能的情况：

int32_t diff = num1 - num2;
bool isZeroOrOne = (diff == 0 || diff == 1 || diff == -1);

有没有更优化的方法？

Answer 1

编译器已经知道像这样的范围比较的优化技术，不需要做任何花哨的事情，只需使用它

int32_t diff = uint32_t(num1) - uint32_t(num2);
bool isZeroOrOne = diff >= -1 && diff <= 1;

(uint32_t)(num1 - num2 + 1) <= 2正如 Raymond Chen 评论的那样，编译器将优化为单个比较

unsigned需要进行强制转换以避免由于整数溢出而导致未定义的行为。-fwrapv或者在支持的编译器中使用

Godbolt 上的演示。正如你所看到的，编译器将下面的函数编译成完全相同的指令

#include <cstdint>

bool diff1(int32_t num1, int32_t num2)
{
    int32_t diff = uint32_t(num1) - uint32_t(num2);
    return diff >= -1 && diff <= 1;
}

bool diff2(int32_t num1, int32_t num2)
{
    int32_t diff = uint32_t(num1) - uint32_t(num2);
    return (uint32_t)(diff - (-1)) <= (1 - (-1));
}