[英]Implementing “logical not” using less than 5 bitwise operators
As part of my CS classes I've recently completed the pretty popular "Data Lab" assignments. 作为我的CS课程的一部分,我最近完成了非常受欢迎的“数据实验室”任务。 In these assignments you are supposed to implement simple binary operations in C with as few operations as possible.
在这些分配中,您应该使用尽可能少的操作在C中实现简单的二进制操作。
For those who are not familiar with the "Data Lab" a quick overview about the rules: 对于那些不熟悉“数据实验室”的人,可以快速了解规则:
The task is to implement a logical not called 'bang' (where bang(x) returns !x) by only using the following operators: ~ & ^ | 任务是通过仅使用以下运算符来实现一个不称为'bang'的逻辑(其中bang(x)返回!x):〜&^ | + << >>
+ << >>
The function prototype is defined as 函数原型定义为
int bang(int x)
The best implementation I could find (using 5 operators) was the following: 我能找到的最佳实现(使用5个运算符)如下:
return ((x | (~x +1)) >> 31) + 1
However there seems to be a way to accomplish this with even less operators, since I found a result website[1] from some German university where two people apparently found a solution with less than 5 operator. 然而,似乎有一种方法可以通过更少的操作员来实现这一目标,因为我在一些德国大学找到了一个结果网站[1],其中两个人显然找到了一个少于5个操作员的解决方案。 But I can't seem to figure out how they accomplished that.
但我似乎无法弄清楚他们是如何完成的。
[1] http://rtsys.informatik.uni-kiel.de/~rt-teach/ss09/v-sysinf2/dlcontest.html (logicalNeg column) [1] http://rtsys.informatik.uni-kiel.de/~rt-teach/ss09/v-sysinf2/dlcontest.html(logicalNeg专栏)
To clarify: This is not about how to solve the issue, but how to solve it with less operations. 澄清:这不是关于如何解决问题,而是如何用较少的操作来解决问题。
Only slightly cheating: 只是轻微作弊:
int bang(int x) {
return ((x ^ 0xffffffffU) + 1UL) >> 32;
}
is the only way I can think of to do it in only 3 operations. 是我能想到的只有3个操作的唯一方法。 Assumes a 32-bit int and 64-bit long...
假设32位int和64位长...
If you take the liberty of assuming that int addition overflow is well-defined and wraps (rather than being undefined behavior), then there's a solution with four operators: 如果您冒昧地假设int加法溢出是明确定义并且包装(而不是未定义的行为),那么有一个包含四个运算符的解决方案:
((a | (a + 0x7fffffff)) >> 31) + 1
I think you are assuming that overflow is defined to wrap otherwise your function ((x | (~x + 1)) >> 31) + 1
has undefined behavior for x=INT_MIN. 我认为你假设溢出被定义为包装,否则你的函数
((x | (~x + 1)) >> 31) + 1
具有x = INT_MIN的未定义行为。
why not just :- 为什么不呢: -
int bang(int x)
{
return 1 >> x;
}
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