c bitwise-operators bit logical-operators

执行逻辑“不”！仅使用按位运算

发布于 2024-09-29 06:31:52 字数 363 浏览 1 评论 0原文

可能的重复：
使用以下命令检查数字是否非零C 中的按位运算符。

大家好，

我正在做一个项目，我需要一些关于函数的帮助。我们需要编写一个函数来执行逻辑非，！，仅使用以下按位运算符：

~ & ^ | + << >>

我什至不知道从哪里开始。

原文

Possible Duplicate:
Check if a number is non zero using bitwise operators in C.

Hello everyone,

I am working on a project and I need a little help with a function. We need to write a function that performs the logical not, !, using only the following bitwise operators:

~ & ^ | + << >>

I'm not even sure where to begin.

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空‖城人不在 2024-10-06 06:31:52

如果您可以假设 true = 1 和 false = 0，那么这可能会起作用：

bool
not(bool x) {
    bool not_x = x ^ true;
    return not_x;
}

If you can assume that true = 1 and false = 0, then this might do the trick:

bool
not(bool x) {
    bool not_x = x ^ true;
    return not_x;
}

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嘿看小鸭子会跑 2024-10-06 06:31:52

如果值不为零，逻辑非返回 0，否则返回 1。假设 32 位 int：

int not_func(int v) {
    /* compress value to single bit */
    int p = (v >> 16) | v;
    p = (p >> 8) | p;
    p = (p >> 4) | p;
    p = (p >> 2) | p;
    p = (p >> 1) | p;

    p ^= 1;
    return (p & 1);
}

Logical not returns 0, if value is not zero, 1 otherwise. Assuming 32-bit int:

int not_func(int v) {
    /* compress value to single bit */
    int p = (v >> 16) | v;
    p = (p >> 8) | p;
    p = (p >> 4) | p;
    p = (p >> 2) | p;
    p = (p >> 1) | p;

    p ^= 1;
    return (p & 1);
}

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故事与诗 2024-10-06 06:31:52

我认为首先你需要澄清这个问题。听起来您想要一个函数，如果单词中的任何位为“1”，则返回 0；如果所有位都为零，则返回 0 以外的值。假设有一个 32 位单词“a”，你可以这样做：

na1 = ~a;
shifted_na1 = na1 >> 1;
na2 = shifted_na1 & na1; /* any sequence of zeros is now 2 bits long */
shifted_na2 = na2 >> 2;
na3 = shifted_na2 & na2; /* any sequence of zeros is now 4 bits long */
shifted_na3 = na3 >> 4;
na4 = shifted_na3 & na3; /* any sequence of zeros is now 8 bits long */
shifted_na4 = na4 >> 8;
na5 = shifted_na4 & na4; /* any sequence of zeros is now 16 bits long */
shifted_na5 = na5 >> 16;
final = shifted_na5 & na5; /* any sequence of zeros is now 32 bits long */
really_final = final & 1;

I think to start you want you'll want to clarify the question. It sounds like you're wanting at a function that will return 0 if any of the bits in a word are "1", and something other than 0 if all the bits are zero. Assuming a 32bit word "a" you could do something like:

na1 = ~a;
shifted_na1 = na1 >> 1;
na2 = shifted_na1 & na1; /* any sequence of zeros is now 2 bits long */
shifted_na2 = na2 >> 2;
na3 = shifted_na2 & na2; /* any sequence of zeros is now 4 bits long */
shifted_na3 = na3 >> 4;
na4 = shifted_na3 & na3; /* any sequence of zeros is now 8 bits long */
shifted_na4 = na4 >> 8;
na5 = shifted_na4 & na4; /* any sequence of zeros is now 16 bits long */
shifted_na5 = na5 >> 16;
final = shifted_na5 & na5; /* any sequence of zeros is now 32 bits long */
really_final = final & 1;

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