Bitwise operation bit check

Question

Is this correct way to check if bits are set in some variable and store them in another variable? Just ignore variable names, image[] and mask variables are 1 byte and first_hline_first_row and second_hline_first_row are 2 byte.

 for(i=0; i<16; i++){

    if(image[i] & mask1)         first_hline_first_row    |=   (1<<i);
    else                         first_hline_first_row    &=~  (0<<i);

    if(image[i] & mask2)         second_hline_first_row   |=   (1<<i);
    else                         second_hline_first_row   &=~  (0<<i);
    }

Shifting zero yield no effect. As well as OR-ing with it. ANDing with it will produce zero as well. — Eugene Sh., May 25 '15 at 14:11
Okay so what is this? http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c-c — Lucky, May 25 '15 at 14:14
Don't you notice a *little* difference between your code and the link? First understand what the code should do, and then use it. — Eugene Sh., May 25 '15 at 14:15
Do it step by step. Think, what should be done with a number to clear a specific bit in it? Right. AND it with a number which is all ones except that bit. How to calculate this number? Right. Take an all-zeros number with rightmost bit `1` and shift it until the one in the correct place. Than invert the bits. — Eugene Sh., May 25 '15 at 14:21
regarding this kind of line, which is trying turn a bit OFF. 'second_hline_first_row &=~ (0< — user3629249, May 25 '15 at 16:09

banach-space · Accepted Answer · 2015-05-25T14:21:59.050

0

Two things:

image[i] & mask1 only checks whether image[i] and mask1 have at least one bit in common. Is this your intention?
first_hline_first_row &=~ (0<<i); does nothing (~(0<<i) is all ones).

From the comments I'm guessing that this is what you meant to write:

for(i=0; i<16; i++){
     if(image[i] & mask1)
     {
         first_hline_first_row    |=   (1<<i);
     }
     else                         
     {
         first_hline_first_row    &=~  (1<<i);
     }

     if(image[i] & mask2)
     {
         second_hline_first_row   |=   (1<<i);
     }
     else
     {
         second_hline_first_row   &=~  (1<<i);
     } 
}

edited May 25 '15 at 14:21

answered May 25 '15 at 14:14

banach-space

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Yes just one bit is checked in that AND, I want to set some bits in new variable with the help of masking. – Lucky May 25 '15 at 14:15
Thank you for pointing me to what was wrong in code, now I understand why it didn't work. – Lucky May 26 '15 at 06:27
I have problem when declaring same for loop which do the same as that code you declared, its range goes from 16 to < 32 and variables where bits are stored are with different names, first for loop works but second wont work, whats the problem? – Lucky May 26 '15 at 11:10
It'll be much easier and quicker if you attach your code. – banach-space May 26 '15 at 11:14
I can't I'm limited for a few days. – Lucky May 26 '15 at 11:17
here is a code http://pastebin.com/6CQJHDEg, just ignore variable names, the problem is that second for loop wont work, DISPLAY_WIDTH is 16 – Lucky May 26 '15 at 11:20
Are `druga_slajfna_prvi_red` and `druga_slajfna_prvi_red` 4 bytes wide? – banach-space May 26 '15 at 11:23
They are both two bytes wide – Lucky May 26 '15 at 11:25
I found where its problem it is trying to shift 16-32 places to left but it cant because variable is 16bits, am I right? All significant bits go out of variable bit range :) – Lucky May 26 '15 at 11:26
We'll here's your answer: `i` ranges from 16 to 31 and so `(1< – banach-space May 26 '15 at 11:28
Correct and it's me who found the problem ;-P – banach-space May 26 '15 at 11:28
Anyway, could you upvote my original answer? Provided you haven't done it yet. Keep warm! – banach-space May 26 '15 at 11:29
I will if I have 15 reputation here on stackoverflow – Lucky May 26 '15 at 11:30

Bitwise operation bit check

1 Answers1