Converting from 8 bits to 1 byte

若如初见. 提交于 2019-12-05 13:05:13
sum += bits[i];

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';

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

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--)

should really be

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

This should work properly, under normal situations.

What does the char[] contain? Does it contain '0', '1' instead of 0, 1?

To make sure, change line

sum += bits[i];

to

sum |= bits[i] & 1;

Also, as pointed out by Vaughn Cato in comments, you need to shift before adding.

If any one is still interested in converting bits to bytes check out this gist. I prepared it as a sketch for the Arduino so that I can send out the electrical pulses that represent each byte. I used arrays of int instead of character but the concept is the same. Considering the size of those types, I may soon switch it to byte or char.

in particular take a look at the encodeByte and decodeByte functions

Sketch for transmitting bits gist

#include <stdint.h>

int main()
{
    uint8_t aa=0x41; 
    printf("%c",aa);  //A
}
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