如何在没有相等运算符的情况下区分两个整数？

Question

Given two integers, how can I differentiate between the two integers, maximum, minimum and equality without using if statements or selections (>=, >, and =). I thought about using fmin and fmax , but I also need to know how to know if the two integers are the same.

Answer 1

Assuming 2's complement arithmetic:

int intcmp(int int1, int int2)
{
    int diff=int2-int1;
    unsigned int udiff;
    memcpy(&udiff, &diff, sizeof(diff));
    if(!udiff)
        return 0;  /* the two integers are equal */
    else if(udiff & 1<<(sizeof(udiff)*CHAR_BIT-1))  /* check the sign */
        return +1;  /* int1 < int2 */
    else
        return -1;  /* int2 < int1 */
}

Answer 2

怎么样...

(num1-num2) ? ((num1-num2) & (MAX_INT+1) ? NEGATIVE : POSITIVE) : EQUAL

Answer 3

What do you mean by "differentiate"? You can't cause two numbers to select two different execution paths without some sort of branch operation, but you could in theory load an array with function pointers and manipulate the numbers to select different array elements to call (and a call is a form of branch).

You could somewhat more easily load an array with two different character string pointers, and select which to print based on which number you get.

You can convert a number to 0 or 1 based on whether it's odd or even by ANDing with 1, eg. You can determine if numbers are equal by subtracting them and seeing if you get zero or not. Etc.

Answer 4

You could use the following to generate a mask

signed int x = ...;
signed int y = ...;
mask = (x - y) >> 31;

So mask is -1 if y>x, and is 0 otherwise (but be careful of overflow). You can then use & , | , and ~ to build your result using this mask.