Unsigned Character in C

Question

unsigned char num;
unsigned char b0=1;
unsigned char b1=0;
unsigned char b2=1;

How to assign num to b2b1b0? If we print num in binary at the end, it should be 101.

Hello, welcome to StackOverflow! Could you please provide with the full code (i do not know where `x` come from) and tell us what you have tried so far? — John Odom, Nov 07 '14 at 21:47
There shouldn't be any x. I just realize << only shifting to the left — Ichihara26, Nov 07 '14 at 22:12
possible duplicate of [How do you set, clear and toggle a single bit in C/C++?](http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c-c) — Alex Celeste, Nov 08 '14 at 00:05

djhaskin987 · Accepted Answer · 2014-11-07T22:38:05.657

0

Assuming b0, b1, and b2 are all bits within the num variable, and assuming little-endianness:

num = 1 | 4;

If you really needed to use the b* variables (I'm assuming you don't), You could do this:

unsigned char b0 = 1U;
unsigned char b1 = 2U;
unsigned char b2 = 4U;
unsigned char num = b0 | b2;

edited Nov 07 '14 at 22:38

answered Nov 07 '14 at 22:28

djhaskin987

Can you explain what's 1 | 4 in c? Thanks – Ichihara26 Nov 07 '14 at 22:33
The number 1 *bitwise-or*'ed with the number 4, or, in binary, '`1` or `100`' -- which equals `101`. – djhaskin987 Nov 07 '14 at 22:36
`1 | 4` indicates a bitwise OR operation between the two values. Written in binary as three bits each, this is `001 | 100`. The OR operation makes a new value, keeping the `1` bits wherever they occur. In this case the result is `101` which is 5 in decimal. Hopefully that makes sense of `1 | 4 = 5`. Coincidentaly the result here is the same as adding them would be, but can you see that `1 | 5 = 5` too? – Weather Vane Nov 07 '14 at 22:40
b* variables are unknown. b* are the LSBs from 3 different random numbers – Ichihara26 Nov 07 '14 at 22:48

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