4

有时我想做这样的事情（i 和 j 是整数）。

(if i==4 && j==9)
{
   ...
}

如果 i 等于 4 且 j 等于 9，它将通过方括号。我一直在使用单与号 (&) 而不是双一，并且我的代码一直在编译和运行。

它是否与双 && 做同样的事情，如果不是，它一直在做什么？

编辑：哦，我一直在用 or 做同样的事情，使用 '|' 而不是'||'

4

4 回答 4

9

Presumably you mean if (i==4 && j==9).

Under the circumstances, changing this from && to & shouldn't change much. The big thing that'll change is that with &&, the j==9 would only be evaluated if the i==4 part was true, but with &, they'll both be evaluated regardless.

When you have something like if (x != NULL && x->whatever ...) you want to ensure that the second part (that dereferences x) is only evaluated if x is not a null pointer. In your case, however, comparing what appear to be ints is unlikely to produce any problems.

It's also possible to run into a problem when you're dealing with something that may produce a value other than 1 to signal true. Again, it's not a problem here because == will always produce either 0 or 1. If (for example) you were using isalpha, islower, etc., from <ctype.h>, they're only required to produce 0 or non-zero values. If you combined those with a &, you'd get a bit-wise or which could produce 0 for two non-zero inputs (e.g., 1 & 2 == 0, but 1 && 2 == 1).

When you use bitwise and on the results from ==, you're going to get 0 & 0 or 0 & 1 or 1 & 0 or 1 & 1. 1 & 1 will yield 1 (true). All the others will yield 0 (false) --- just like && would have.

于 2012-12-19T02:25:49.230 回答
5

它正在执行按位 AND。

它所做的是表达式i == 4等价于1if iis4并且对于 RHS 相同（显然带有jand 9）。&运算符返回两个操作数都有该位的数字。

它的工作原理与&&因为00000001 & 00000001（例如缩小到一个字节）相同。如果一个是0（条件为假），则&不会看到两个操作数都打开了两个位，并且00000000is 0，这是falsey。

但是，不要简单地使用&，因为它比一个字符短或相似。使用它，因为它表达了您想要实现的目标。如果它是您想要的逻辑 AND，请使用&&.

同样的事情适用于|，除了如果每个操作数都打开它而不是结果位，它会在任一操作数打开该位时打开它。

于 2012-12-19T02:21:08.583 回答
2

单个 & 符号执行按位与。仅当两个操作数在该位置都为 1 时，才会设置结果的每一位。

Since comparisons in C return 1 for true and 0 for false, & will give the same results as && as long as both operands are comparisons. But for arbitrary values, it will return seemingly random results. 1 && 2 is true, but 1 & 2 is false because the binary representations of 1 and 2 have no bits in common.

于 2012-12-19T02:22:20.240 回答
0

单个＆符号称为按位与。它是一个二进制运算符，逐位“与”两个数字。

例如，如果你有两个二进制数，01100010 和 00011111，你的按位和将导致 00000010

i==4 返回 1（真），j==9 也是如此。因此，i==4 & j==9 实际上只是 1 & 1，其计算结果为 1（真）。

试试这些例子看看有什么不同

1) 整数 k = 4 & 6; vs int k = 4 && 6;

2) if(i==4 && 2) 与 if(i==4 & 2)

于 2012-12-19T02:21:33.297 回答