[WildCard65]

How, I can't find any documentation on them.
Edit: Thx to people who help clarify to me on bitstrings and for showing me indirectly my mistake.

[Powerlord]

Quick rundown of bitstrings:

The major operators for bitstrings are:

| // bitwise OR, returns a combination of bits that are set on both sides (used to combine bits into a single bitstring)
& // bitwise AND, returns all bits that exist on both sides (used to check if particular bits are set)
^ // bitwise XOR, reverses all bits that exist on both sides, returns all other bits unchanged
~ // bitwise NOT, unary operator that flips all the bits of its argument
<< // arithmetic left shift, makes all bits X bits larger, lower bits are set to zero
>> // arithmetic right shift, makes all bits X bits smaller, higher bits are set to the sign bit (this is usually what you want)
>>> // logical right shift, makes all bits X bits smaller, higher bits are set to zero

You can add bits to a bitstring using | or |=, such as this:

PHP Code:

new bits = 0;
bits |= (1 << 0); 


You can check if a bit is set using &.

PHP Code:

if (bits & (1 << 0))
{
    // Do something if bit 1 is set
} 


Note that checking multiple bits means you have to use the long form with ==

PHP Code:

if (bits & (1 << 0| 1 << 1) == (1 << 0|1 << 1))
{
    // Do something if bits 1 and 2 are set
} 


You can remove a bit using &= and ~

PHP Code:

bits &= ~(1 << 0); // removes bit 1
// This works because ~(1 << 0) means "every bit except 1" 


[WildCard65]

K thx

[Bacardi]

Originally Posted by Powerlord Note that checking multiple bits means you have to use the long form with == PHP Code: if (bits & (1|2) == (1|2)) {     // Do something if bits 1 and 2 are set }

How This work exactly ? :/
*
(1|2) == (1|2) ??

[h3bus]

& (And operator) has highest precedence

This is interpreted as:

(bits & (1|2)) == (1|2))

[Mitchell]

Originally Posted by Bacardi How This work exactly ? :/ * (1|2) == (1|2) ??

(1|2|4) == (1|2) would return false, it first checks to see if both of those are set in the bits, then checks to see if that's the only bits in the bitstring. That's my guess, could be wrong.
as it would be translated to:
7 == 3

[Doodil]

correct me if I'm wrong, but in other programing languages (1|2|3) does not mean "set bit 1, 2 and 3" but "set the bits that are used to represent the numbers 1, 2 and 3". And that means:

1= 001
2= 010
3= 011

and going with this logic (1|2|3) = 011 and (1|2) = 011, so (1|2|3) == (1|2) equals true.


If you want to set bit number X (starting at 0) then you have to left-shift the first bit by X:

(1<<0) = 001 //equals 1
(1<<1) = 010 //equals 2
(1<<2) = 100 //equals 4, not 3



Please correct me if this is any different in sourcemod.

[Mitchell]

Originally Posted by Doodil correct me if I'm wrong, but in other programing languages (1|2|3)

you caught me: i meant "(1|2|4)"

[Bacardi]

oh darn, now I get it, some how... thanks for explaining
that if statement is like

PHP Code:

if( 0 < client <= MaxClients ) 


*edit wait a second... need think this later

[Mitchell]

Originally Posted by Bacardi oh darn, now I get it, some how... thanks for explaining that if statement is like PHP Code: if( 0 < client <= MaxClients )  *edit wait a second... need think this later

It's helpful, you also learn this in subnet masking, i believe, i cant remember if that is what it's called.
http://en.wikipedia.org/wiki/Logical_bitwise_operator