C中的按位串联 [英] Bitwise concatenation in C

问题描述

我试图在C中连接两个二进制数.因此，如果我有1010和0011，我希望结果为10100011.我写了一个简短的例程，认为可以完成这项工作:

I'm trying to concatenate two binary numbers in C. So if I have 1010 and 0011 I want my result to be 10100011. I wrote a short routine that I thought would do the job:

#include <stdio.h>

int main(void) {
    int first = 1010;
    int second = 0011;
    int result = (first << 4) | second;
    printf("%d", result);
    return 0;
}

我知道打印的数字当然是十进制的，但是我发现按位运算后，我得到的十进制等效值为10100011或163.但是，我的结果打印为16169 .所以我想我的问题是...在这里我不理解其中的哪一部分?这仅仅是对printf的工作方式的误解，还是我的按位操作不正确?尝试使用int s执行此操作是否有问题?

I understand that the printed number of course is going to be in decimal, but I figured that after my bitwise operations I'd be getting the decimal equivalent of 10100011, or 163. However, my result is printed as 16169. So I guess my question is...what part of this am I not understanding here? Is it just a misunderstanding of how printf is working, or are my bitwise operations incorrect? Is it a problem to try to do this with ints?

推荐答案

您忘记了前缀0b，因此它应该对您有用:

You forgot the prefix 0b, so this should work for you:

#include <stdio.h>

int main() {

    int first = 0b1010;
              //^^v See here
    int second = 0b0011;
    int result = (first << 4) | second;
    printf("%d", result);

    return 0;

}

输出:

163

在您的示例中，二进制数"不是二进制数.第一个是普通的十进制数(1010)，第二个是八进制数，因为有前缀0，所以在十进制中，第二个数字是:9

In your examle the 'binary numbers' aren't binary numbers. The first one is a normal decimal number (1010) and the second one is a octal number, because of the prefix 0 so in decimal the second number was: 9

所以发生了什么事

1010 -> decimal
0011 -> octal

第一个电话号码

First number:

     11 1111 0010      
----------------- << 4
11 1111 0010 0000

第二个数字(->十进制->二进制):

Second number (->decimal->binary):

  octal    decimal       binary
  0011  ->    9     ->  0000 1001

以及您的计算:

And your calculation:

11 1111 0010 0000   first number
             1001   second number
----------------- |
11 1111 0010 1001 = 16169

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