将2 char转换为1 int [英] Convert 2 char into 1 int

问题描述

我有两个字符：HIGH和LOW，我想将它们转换为一个int对应于HIGH +从左边的2个左位。

I have 2 chars: HIGH and LOW and I'd like to convert them to an int corresponding to HIGH + the 2 left bits from LOW.

有人喜欢：

unsigned char high;
unsigned char low;
high = 128; // 10000000
low= 128; // 10000000
int result; (should be high 10000000 + 2 left bites of low 10 = 1000000010)
// To do
return result;

编辑为了更清楚。

我选择的解决方案是：

return high*4 + (low >> (CHAR_BIT - 2));

推荐答案

您声明 HIGH 和 LOW 作为 char * ，但不要将它们用作指针。下面的代码工作正常（BTW，避免大写标识符，当你不使用常量）：

You declare HIGH and LOW as char*, but you don't use them as pointer. The following code works fine (BTW, avoid upper case identifiers when you don't use constants):

char high = 125;
char low = 12;

这是我的理解你的问题（它可以更容易理解）：

This is how I understand your question (it could be more understandable):

#include <limits.h>

int result = high + (low >> (CHAR_BIT - 2));

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