整数减法回绕N位

本文介绍了整数减法回绕N位的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

基本上，溢流用减法整数时，但对于位的给定数目的你的行为。最显而易见的方法，假设签署整数：

Basically, the behavior you get when overflowing integers with subtraction, but for a given number of bits. The obvious way, assuming a signed integer:

template <int BITS> int sub_wrap(int v, int s) { int max = (1<<(BITS)); v -= s; if (v < -max) v += max*2; // or if branching is bad, something like: // v += (max*2) * (v < -max) return v; } // For example subtracting 24 from -16 with 5 bit wrap, // with a range of -32, 31 sub_wrap<5>(-16, 28); -> 20

是否有这样做是不太丑陋和$ P $比上面的人？

Is there a neat way of doing it that is less ugly and preferably faster than the one above?

更新：很抱歉的混乱。我不假思索地使用包含排除的叹息位的位数的混乱符号。所以在上面，更换6位5位为更大量的理智。

UPDATE: Sorry about the confusion. I thoughtlessly included the confusing notation of using the number of bits excluding the sigh bit. So in the above, replace 5 bits with 6 bits for a lot more sanity.

推荐答案

有关无符号运算，并掩盖的结果，例如：

For unsigned arithmetic, and mask the results, e.g.:

template<int bits> unsigned sub_wrap( unsigned v, unsigned s ) { return (v - s) & ((1 << bits) - 1); }

更一般地，你可以使用模运算符：

More generally, you can use the modulo operator:

template<int modulo> unsigned sub_wrap( unsigned v, unsigned s ) { return (v - s) % modulo; }

（包装纸上的 N 位是模相当于2 ^ N）。

(Wrapping on n bits is the equivalent of modulo 2^n.)

有关签署算术，它更复杂一点;使用面膜，你必须符号扩展的结果（假设2的补数）。

For signed arithmetic, it's a bit more complex; using the mask, you'll have to sign extend the results (supposing 2's complement).

编辑：使用sehe的建议有符号的算术：

Using sehe's suggestion for signed arithmetic:

template<int bits> int sub_wrap( int v, int s ) { struct Bits { signed int r: bits; } tmp; tmp.r = v - s; return tmp.r; }

鉴于此， sub_wrap小于5＆GT;（-16，28）给 -12 （这是正确的＆MDASH;注意， 28 不能重新$ p $ 5位psented作为符号int）; sub_wrap 6;方式＆gt;（-16，28）给 20

Given this, sub_wrap<5>( -16, 28 ) gives -12 (which is correct—note that 28 cannot be represented as signed int in 5 bits); sub_wrap<6>( -16, 28 ) gives 20.