Mask by n bytes

Question

I struggle masking a uint64_t variable by N bytes. I do not know N but I know it is 8 or less. My current code looks like this:

// uint64_t n is given
uint64_t mask;
for( mask = 0x00; n; n--) {
    mask = mask<<8 | 0xFF;
}

for building the mask. What am I doing wrong here?

Edit:
The question was answered. Anyway, for better understanding:

I want a mask like this:

0x000000FF // or:
0x0000FFFF // or:
0x00FFFFFF

to take 1, 2 or more byte from the data. As the comments say, my code works! Maybe I had an bug anywere else!

Works fine for me. Is this the real code. Explain what is wrong with it. — Pete, Jan 24 '13 at 10:40
An example of input data and expected output data would also help us to provide a better answer — Andreas Fester, Jan 24 '13 at 10:53

unwind · Accepted Answer · 2013-01-24T10:45:25.187

2

It should work, according to the [operator precendence table1].

Still, it's more clear to write it as:

mask <<= 8;
mask |= 0xff;

or:

mask = (mask << 8) | 0xff;

You can also do it with a look-up table, of course.

edited Jan 24 '13 at 10:45

answered Jan 24 '13 at 10:37

unwind

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No he's not - << takes precedence over |. (you did read that link yourself?). It would only fail if he used + instead of | – Pete Jan 24 '13 at 10:39
Actually I retract the -1 because adding the brackets is still better coding style. – Pete Jan 24 '13 at 10:39
@Pete Thanks, I was actually a bit uncertain but didn't take the time to test. I tried to express that, but not very clearly. Rewritten. – unwind Jan 24 '13 at 10:45

Andreas Fester · Answer 2 · 2013-01-24T10:53:54.360

1

I am not sure if I get the question right, but your mask looks like

0x00000000000000ff
0x000000000000ffff
0x0000000000ffffff
...

I assume that you want something like the following, to mask individual bytes:

0x00000000000000ff
0x000000000000ff00
0x0000000000ff0000
...

For this, you can use e.g. the following code:

for( mask = 0xff; n; n--) {

    // Use the mask HERE
    ...

    mask = mask<<8;
}

edited Jan 24 '13 at 10:53

answered Jan 24 '13 at 10:47

Andreas Fester

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1

Well, that's my fault. My question was badly worded, I wanted the first one. Anyway, you are right! – musicmatze Jan 24 '13 at 10:58

score 0 · Answer 3 · answered Jan 24 '13 at 10:49

You could use this code snippet, to replace the bytenoth byte with dest mask, in src:

uint64_t replacemyByte(uint64_t src, uint64_t byteno,uint64_t dest) 
{
    uint64_t shift = (dest << (8 * byteno));
    uint64_t mask = 0xff << shift;
    return (~mask & src) | shift;
}

Or did I get the question wrong?

Mask by n bytes

3 Answers3