用C逻辑运算符 [英] Logical Operators in C

问题描述

我有麻烦试图了解逻辑运算符在C.工作我已经明白是怎么位级运营商的工作，我也知道，逻辑运算符把非零参数如重新presenting TRUE和零参数作为重presenting FALSE

I am having trouble trying to understand how logical operators work in C. I already understand how the bit-level operators work, and I also know that logical operators treat nonzero arguments as representing TRUE and zero arguments as representing FALSE

不过，说我们有0x65和放大器;＆安培;将0x55。我不明白为什么这怎么操作提供了为0x01。

But say we have 0x65 && 0x55. I do not understand why and how this operations gives 0x01.

我试着将它转换为二进制，但我无法弄清楚它是如何工作

I tried to convert it to binary, but I cannot figure out how it works

推荐答案

的＆放大器;＆安培; 是一个逻辑和（而不是＆安培; ，这是一个的按位的和）。它关心只有它的操作数作为零/非零值。零被认为是假，而非零被视为真正。

The && is a logical AND (as opposed to &, which is a bitwise AND). It cares only that its operands as zero/non-zero values. Zeros are considered false, while non-zeros are treated as true.

在您的情况下，两个操作数不为零，因此，他们都以真正，导致的结果是真正为好。 c重新presents 真正为 1 ，解释你的操作的整体效果。

In your case, both operands are non-zero, hence they are treated as true, resulting in a result that is true as well. C represents true as 1, explaining the overall result of your operation.

如果您更改操作＆安培; ，你会得到一个按位运算。 0x65和放大器;将0x55 会给你×45 的结果。

If you change the operation to &, you would get a bitwise operation. 0x65 & 0x55 will give you a result of 0x45.

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