从8位转换为1个字节 [英] Converting from 8 bits to 1 byte

问题描述

我有一个8位的字符串，我想将其转换为1个字节.我不确定为什么我的功能不能正常工作.我有8位存储到8个未签名字符的数组中.到目前为止，这是我的方法:

I have a string of 8 bits and I want to convert it into 1 byte. I am not sure why my function is not working properly. I have 8 bits stored into an array of 8 unsigned chars. This is my method so far:

unsigned int bitsToBytes(unsigned char *bits)
{
  unsigned int sum = 0;
  for(int i = 7; i >= 0; i--)
  {
    sum += bits[i];
    sum<<=1;
  }
  return sum;

}

int main()
{
  unsigned char bits[8];
  unsigned int byt;
  byt = bitsToBytes(bits);
  cout << byt; //doesn't give me the right result
}

我的位数组在数组中包含"1"和"0"！抱歉，不清楚.

My array of bits contains '1' and '0' in the array! Sorry for not being clear.

也许有人知道我在哪里出错了吗?我不确定为什么我的位没有正确转换为字节.有人可以帮忙吗?谢谢！

Might anyone know where I went wrong in this? I'm not sure why my bits aren't converting to bytes properly. Could anyone help? Thanks!

推荐答案

sum += bits[i];

如果您尝试转换C 字符串(例如，"1010101")，则此代码会添加char的代码表值(ASCII，UTF-8，无论您使用哪种编码) (例如48和49)，而不是1和0.您应该将其重写为

If you're trying to convert a C string (for example, "1010101"), this code adds the codetable value (ASCII, UTF-8, whichever encoding you have) of the char (for example, 48 and 49), not 1 and 0. You should rewrite this as

sum += bits[i] - '0';

此外，您不会初始化bits数组-在初始化之前使用其内容会导致未定义的行为，因此您可以弄清任何事情发生.

Also, you don't initialize the bits array - using its contents before initialization results in undefined behavior, so you can expact anything to happen.

此外，您的代码逻辑有缺陷-一种，您必须在添加二进制数字之前向左移动.第二，您正在向后遍历字符串；线

Furthermore, your code logic is flawed - One, you have to do the left shift before adding the binary digit. Two, you're traversing the string backwards; the line

for (int i = 7; i >= 0; i--)

应该是

for (int i = 0; i < 8; i++)

这篇关于从8位转换为1个字节的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！