Can anyone explain me this line? c |= 1 << i;

Question

I recently started C and for some reasons I couldn't get this line c |= 1 << i;

the purpose of this function I found online is to get the least significant bit from an array and then combine it, and return as a byte.

unsigned char getlsbs(unsigned char* p)
{
        int i;
        unsigned char c = 0;
        for(i = 0; i < 8; i++)
        {
                int a = p[i] & 1;
                if(a)
                {
                        c |= 1 << i;
                }
        }
        return c;
}

c |= 1 << i; would be the same as c = c | 1 << i; correct?

Could anyone explain with the example in 1s and 0s? I think it will be very helpful. Thanks!

Answer 1

Well,

1<<i 

should be 1 followed by i zeros (binary)--so

1<<0 = 0001
1<<1 = 0010
1<<2 = 0100

When that is ORed with what's in C it means to force-set that bit, so:

if you take 0000 | 0010 you'll get 0010

c val| mask = result
--------------------
0010 | 0010 = 0010 as well
1111 | 0010 = 1111 (No change if the bit is already 1)
1101 | 0010 = 1111 (Just sets that one bit)
xxxx | 0010 = xx1x (All the other bits remain the same)

Finally it stores the result back into c.

So essentially it sets the ith bit in c (starting from the least significant as zero).

Further detail:

// Loop through the first 8 characters in p
for(i = 0; i < 8; i++)
{
    // Grab the least significant bit of the i'th character as an int (1 or 0)
    int a = p[i] & 1;
    // If it happens to be set (was a 1)
    if(a)
    {
        // Set the corresponding (i'th) bit in c
        c |= 1 << i;
    }
}

So if the first value of p has a lsb of 1, the lsb of c will be 1

if the second byte in p has a lsb of 1, the second bit of c will be 1

etc.

the result is that each bit of C will take the value of least significant bit of each of the first 8 bytes of P

I bet it could be coded more densely though if I really wanted to try, but this is probably targeting performance :)