Check multiple bits in bitset

Question

I got the following code:

p = B[m] & B[m + 5] & B[m + 6] & B[m + 11];

m -= d * (l > 0) * 11 + !d * (c % 5 > 0);

p += m ^ M ? B[m] & B[m + 5] & B[m + 6] & B[m + 11] : 0;

I know it's hard to read, but here's a TL;DR for it : I check multiple bits (all are related to m) in a bitset, then i change the value of variable m and i check again (other bits). Is there a way i can acces those bits in less code, or to template the check (cuz are the same formulas for bits)?

B[m] & B[m + 5] & B[m + 6] & B[m + 11]

Thank you :D.

Interestingly enough, your bitset fits into one SSE register, which would allow to precompute the `B[m]&B[m+5]&...` extremely fast for all `m` values at the same time — Ap31, Apr 17 '17 at 18:24
Idk what you talking about, but i don't want to change the structure(bitset) into something else that would ruine the whole code. Could you link me something? — aleeN1, Apr 17 '17 at 18:34
@aleeN1 I don't think you have to change anything, chances are the compiler will make use of it. Just notice that `bitset<99>` is very lightweight so it's perfectly okay to copy it around — Ap31, Apr 17 '17 at 18:37

score 3 · Answer 1 · edited May 23 '17 at 12:17

3

Make a function that takes B and m.

So p = yourFunc(B, m) and p += m ^M ? yourFunc(B, m) : 0

The function is something like:

TYPEOFP yourFunc(TYPEOFB b, TYPEOFM m) {
    return b[m] & b[m + 5] & b[m + 6] & b[m + 11];
}

I don't know your types, so you need to fill it in.

I wouldn't recommend a macro, but if you want that it's

#define yourMACRO(b, m) ((b)[(m)] & (b)[(m) + 5] & (b)[(m) + 6] & (b)[(m) + 11])

All of those extra parens are to protect you if you ever pass in an expression for b or m. The macro will fail if you pass in something with side-effects (like ++m).

EDIT: From your comments, you said you can't write outside the function.

It's unorthodox, but you can do the #define in the function and #undef it at the end of the function.

Depending on the version of C++ you have, you might have lambdas, which let you make function expressions.

If you are desperate, you can define an inner class or struct with a static function: C++ can we have functions inside functions?

edited May 23 '17 at 12:17

Community

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answered Apr 17 '17 at 18:10

Lou Franco

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The code it's from a function. I can't write anything outside of the function. I forgot to specify this. Sorry. – aleeN1 Apr 17 '17 at 18:11
What version of C++? – Lou Franco Apr 17 '17 at 18:13
Latest one. 14 i think it is – aleeN1 Apr 17 '17 at 18:14
Use a lambda: http://stackoverflow.com/questions/7627098/what-is-a-lambda-expression-in-c11 – Lou Franco Apr 17 '17 at 18:16
I thought about it, but i can't figure out how to do it exactly, cuz it's a complex one.. – aleeN1 Apr 17 '17 at 18:21

Ap31 · Accepted Answer · 2017-04-17T21:05:51.893

3

I suggest using a function to precompute a helper bitset for that:

bitset<99> prepare_bitset(const bitset<99>& B)
{
   return B & (B<<5) & (B<<6) & (B<<11);
}

Then you can just use it like this:

auto HB = prepare_bitset(B);
p = HB[m];
m -= d * (l > 0) * 11 + !d * (c % 5 > 0);
p += m ^ M ? HB[m] : 0;

UPD: Another option is to just define HB in place:

auto HB = B & (B<<5) & (B<<6) & (B<<11);
p = HB[m];
m -= d * (l > 0) * 11 + !d * (c % 5 > 0);
p += m ^ M ? HB[m] : 0;

edited Apr 17 '17 at 21:05

answered Apr 17 '17 at 18:32

Ap31

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can i do that inside the function? I'm talking about prepare_bitset function, can i write in inside another function? – aleeN1 Apr 17 '17 at 18:36
I'll try it out. I thought lambdas can only be used as parameters for functions. – aleeN1 Apr 17 '17 at 18:41

Check multiple bits in bitset

2 Answers2