什么是C中的逻辑运算符的意义呢？ [英] What is the point of the logical operators in C?

问题描述

我只是想知道是否有在C异或逻辑运算符（像＆功放;＆安培;对于要不是XOR）。我知道我可以拆分成XOR与运算，穷人和ORS，但一个简单的XOR会好得多。然后，它发生，我认为如果我用两个条件之间的正常XOR位运算符，它可能只是工作。而对于我的测试中，它做到了。

I was just wondering if there is an XOR logical operator in C (something like && for AND but for XOR). I know I can split an XOR into ANDs, NOTs and ORs but a simple XOR would be much better. Then it occurred to me that if I use the normal XOR bitwise operator between two conditions, it might just work. And for my tests it did.

考虑：

int i = 3;
int j = 7;
int k = 8;

就为了这个相当愚蠢的例子来说，如果我需要的 K 的是不是比的 I 的大于或更高的Ĵ的但不两者，XOR将是相当方便的。

Just for the sake of this rather stupid example, if I need k to be either greater than i or greater than j but not both, XOR would be quite handy.

if ((k > i) XOR (k > j))
   printf("Valid");
else
   printf("Invalid");

或

printf("%s",((k > i) XOR (k > j)) ? "Valid" : "Invalid");

我把按位异或的 ^ 的，它生产的无效。将两个比较的结果在两个整数导致2整数含有1，因此，异或产生的错误的。我再与＆放试了一下;和|位运算符和俩给了预期的效果。这一切都有道理知道真实情况有一个非零值，而假的条件是零值。

I put the bitwise XOR ^ and it produced "Invalid". Putting the results of the two comparisons in two integers resulted in the 2 integers to contain a 1, hence the XOR produced a false. I've then tried it with the & and | bitwise operators and both gave the expected results. All this makes sense knowing that true conditions have a non zero value, whilst false conditions have zero values.

我想知道，有没有理由使用逻辑放大器;＆安培;和||当位运算符＆放;, |和^工作是一样的？

I was wondering, is there a reason to use the logical && and || when the bitwise operators &, | and ^ work just the same?

推荐答案

您不需要逻辑XOR，我已经忘记了SO问题，但它是类似于你在想什么，基本上我们不需要XOR，它等同于！= 反正

You don't need logical XOR, I have forgotten the SO question, but it's similar to what you're thinking, basically we don't need XOR, it's equivalent to != anyway

FALSE XOR FALSE == FALSE
FALSE XOR TRUE == TRUE
TRUE XOR FALSE == TRUE
TRUE XOR TRUE == FALSE


FALSE != FALSE == FALSE
FALSE != TRUE == TRUE
TRUE != FALSE == TRUE
TRUE != TRUE == FALSE

我要搜索我的最爱，并粘贴在这里的链接后...

I'll search my favorites, and paste here the link later...

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