C中没有减号的减法 [英] Subtraction without minus sign in C

问题描述

如何在没有-运算符的情况下在C中减去两个整数?

How can I subtract two integers in C without the - operator?

推荐答案

int a = 34;
int b = 50;

您可以使用取反并加1将b转换为负值.

You can convert b to negative value using negation and adding 1:

int c = a + (~b + 1);

printf("%d\n", c);

-16

这是二进制补码负号.当您要否定值或对其进行子跟踪时，如果您使用'-'运算符，则处理器正在执行此操作.

This is two's complement sign negation. Processor is doing it when you use '-' operator when you want to negate value or subtrackt it.

转换浮点数更简单.只需否定第一位(shoosh为您提供了如何执行此操作的示例).

Converting float is simpler. Just negate first bit (shoosh gave you example how to do this).

好的，伙计们.我放弃.这是我的独立于编译器的版本:

Ok, guys. I give up. Here is my compiler independent version:

#include <stdio.h>

unsigned int adder(unsigned int a, unsigned int b) {
    unsigned int loop = 1;
    unsigned int sum  = 0;
    unsigned int ai, bi, ci;

    while (loop) {
        ai = a & loop;
        bi = b & loop;
        ci = sum & loop;
        sum = sum ^ ai ^ bi;      // add i-th bit of a and b, and add carry bit stored in sum i-th bit
        loop = loop << 1;
        if ((ai&bi)|(ci&ai)|(ci&bi)) sum = sum^loop; // add carry bit
    }

    return sum;
}

unsigned int sub(unsigned int a, unsigned int b) {
    return adder(a, adder(~b, 1));    // add negation + 1 (two's complement here)
}


int main() {
    unsigned int a = 35;
    unsigned int b = 40;

    printf("%u - %u = %d\n", a, b, sub(a, b)); // printf function isn't compiler independent here

    return 0;
}

我正在使用unsigned int，以便任何编译器都将其视为相同.

I'm using unsigned int so that any compiler will treat it the same.

如果要减去负值，则可以这样做:

If you want to subtract negative values, then do it that way:

 unsgined int negative15 = adder(~15, 1);

现在，我们完全独立于有符号值的约定.在我的方法中，所有整数都将被存储为二进制补码-因此，对于较大的整数(必须从0位开始)，您必须要小心.

Now we are completly independent of signed values conventions. In my approach result all ints will be stored as two's complement - so you have to be careful with bigger ints (they have to start with 0 bit).

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