setting bits of a uchar

Question

I am trying to set a uchar as follows:

uchar num = 0;
    //0
    num <<= 1;
    //1
    num |=1;
    num <<=1;
    //0
    num <<=1;
    //1
    num |=1;
    num <<=1;
    //0
    num <<=1;
    //1
    num |=1;
    num <<=1;
    //0
    num <<= 1;
    //0
    num <<=1;
    //should be 01010100 = 84
    std::cout << " num is " << num << " int " << (unsigned int) num << std::endl;

Which should end up with the binary 8-bit sequence 01010100 which is 84 in decimal. However, when I print the output I get is num is ® int 168

What am I doing wrong?

Thanks

Answer 1

The last shift is too much. It multiplies 84 by 2 which gives 168

Answer 2

You're shifting for a zero once too often at the end of your series:

//1
num |=1;
num <<=1;
//0
num <<= 1;
//0
num <<=1;

This won't generate a binary 100 but rather 1000 thus your end-value will be

1010 1000 (168)

instead of

1010 100 (84)

Remove one of those last 0 shifts and you'll be good to go: Example

Answer 3

When you want to "append" a 1 , you should first shift, then OR the 1 . Otherwise you shift the just-appended 1 which becomes a 2 , hence your result is exactly twice the number you expected.

//1
num <<= 1;
num |= 1;

//0
num <<= 1;

Then, your last shift is correct (in contrast to the other answers, which are also correct but if you simply remove the last shift your code comments are wrong).