问题描述
我真的已经很努力在C int变量翻转位。我这样做，像这样：
I am trying REALLY hard to flip the bits in a C int variable. I am doing it like so:
input = 15;
input = ~input;
printf("%d", input);
但它总是显示为 -16 。它应该是 0 ！如果 15 写成 1111 ，为什么它返回 10000 ？这是令人发狂！有人可以帮我！？
BUT IT ALWAYS SHOWS UP as -16. It should be 0! if 15 is written as 1111, why is it returning 10000?! This is maddening! Can somebody PLEASE help me!?
推荐答案
由于 INT 您的系统上是最有可能的一个32位的数字，所有位翻转，其中包括那名猥琐的人在零原来的号码：
Since int on your system is most likely a 32-bit number, all bits are flipped, including the ones that were insignificant zeros in the original number:
00000000000000000000000000001111
变为
11111111111111111111111111110000
这是一个负数：最显著位 15 为零，因此它成为翻转时 1
This is a negative number: the most significant bit of 15 is zero, so it becomes 1 when flipped.
如果您想保留原来的号码只有位，你需要在号码的显著位置全部为掩盖，就像这样：
If you would like to keep only the bits of the original number, you need to mask with all ones in the significant positions of the number, like this:
printf("%d\n", input & 0xF);
和 ING与 0xF的切断，除了过去四年所有位。
ANDing with 0xF "cuts off" all bits except the last four.
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