C ++位运算 [英] C++ bitwise operations

问题描述

我在操作两个整数的各个位，（我用G ++为Ubuntu机器上编译）。

I am operating on individual bits of two integers, (i am using g++ for compilation on Ubuntu machine).

在一些中间步骤，我有位再presentations为

In some intermediate step, I have the bit representations as

q = 11000000000000000000000000000000
q_1 = 00000000000000000000000000000001

现在我要检查q和Q_1单位的地方是否都相同与否。所以，我检查（* Q）及1 == Q_1 中的if条件，其做工精细

Now I want to check whether unit's places of q and q_1 are both same or not. so, I am checking (*q)&1==q_1 in the if condition, and its working fine.

但是，每当我要检查该单位的地方问是0和Q_1是1，我想我应该做的（（* Q）及1 == 0）及和放大器; （Q_1 == 1），但它不工作了预期。为了调试，我 COUT ED的值（（* Q）及1 == 0）和（Q_1 == 1）单独和他们得到了打印为 1 。但是，值（（* Q）及1 == 0）及和放大器; （Q_1 == 1）印刷得到了为0。为什么？

But whenever I want to check that unit's place of q is 0 and that of q_1 is 1, I thought I should do ((*q)&1==0) && (q_1==1), but it is not working out as expected. For debugging, I cout ed the values of ((*q)&1==0) and (q_1==1) individually and they got printed as 1. However, the value of ((*q)&1==0) && (q_1==1) got printed as 0. Why?

*的编辑：* 的在功能方面，Q是通过引用传递，所以我使用* Q来获取值。

*EDIT : * In the function, q was passed by reference, so I am using *q to get the value..

推荐答案

在C和C ++中，按位＆安培; 运营商实际上具有较低的precedence比等价运营商 == 。你需要换你位运算符在括号中。

In C and C++, the bitwise & operator actually has lower precedence than the equivalence operator ==. You'll need to wrap your bitwise operators in parentheses.

所以：

((*q)&1==0) && (q_1==1)

应该是：

(((*q)&1)==0) && (q_1==1)

请参阅：<一href=\"http://en.cp$p$pference.com/w/cpp/language/operator_$p$pcedence\">http://en.cp$p$pference.com/w/cpp/language/operator_$p$pcedence

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