[Bugsy]

claudiuhks: That post is over three years old.

[vitorrossi]

Stock that returns the number of SetBits in a bitfield, aka hamming weight or popcount:

PHP Code:

     NumberOfSetBits(iBitField)
    {
        iBitField = iBitField - ((iBitField >> 1) & 0x55555555);
        iBitField = (iBitField & 0x33333333) + ((iBitField >> 2) & 0x33333333);
        return (((iBitField + (iBitField >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
    } 


Taken from here

[PartialCloning]

If I have a variable equal to (1 << value) is it possible to get the value part of it?

[klippy]

I can't think of any efficient way, but what exactly are you trying to achieve? Having to find that value doesn't seem like a good thing to me.

EDIT:
This could work, but I don't like that it's a loop:

PHP Code:

getValue(variable) {
    for(new i = 0; i < 32; i++) {
        if(variable == (1 << i)) {
            return i;
        }
    }
    
    // Some invalid state I guess, variable has more than 1 bit set.
    return -1;
} 


[Emp`]

Alternative to KliPPy's:

PHP Code:

stock BitShiftCount( x )
{
    new i;
    while ( x && !( x & 1 ) )
    {
        x >>= 1;
        i++;
    }
    return i;
} 


[Bos93]

int i

[Depresie]

Now this, this is, this is a fucking bad ass tutorial, these little fuckers can be extremely useful
I tried to learn about bits some time ago, but i couldn't understand even the basic things from what i found on google, but this man is golden when it comes to tutorials
Thanks Bugs

[eyal282]

Can you do for example:

#define SetUserBool(%1,%2) ( %1 |= (1 << (%2 - 1)))

Because it will range from 0 to 31.

[Bugsy]

SetUserBool(%1,%2) ( %1 |= (1 << (%2 - 1)))

You used |= instead of left shift and you should shift 1, not %1, but yea that will work as long as only 1 to 32 are inputted. Same logic as using ( %2 & 31 ).

[eyal282]

Originally Posted by Bugsy SetUserBool(%1,%2) ( %1 |= (1 << (%2 - 1))) You used |= instead of left shift and you should shift 1, not %1, but yea that will work as long as only 1 to 32 are inputted. Same logic as using ( %2 & 31 ).

Yeah, mistake.