[Napoleon_be]

Originally Posted by Bugsy Depends on the value stored. There are 8 bits in a byte, if the value of the byte is 0 the bits will be 0000 0000. If the byte holds 1, the bits will be 0000 0001. 0 = 0000 0000 1 = 0000 0001 2 = 0000 0010 3 = 0000 0011 4 = 0000 0100 5 = 0000 0101 6 = 0000 0110

How about 7? will that be 0000 0111 ?
and 8 would be 0000 1000?

[Bugsy]

yup

[Napoleon_be]

Awesome, thanks guys

[Bugsy]

Here's a mini tutorial for weapons:

new g_WeaponList = ( 1 << CSW_AK47 ) | ( 1 << CSW_M4A1 ) | ( 1 << CSW_AWP )

PHP Code:

#define CSW_AWP             18
#define CSW_M4A1            22
#define CSW_AK47            28 


Step 1 for the bitsum is to take 1 and shift it left the number of slots for the weapon ID.

For AWP, take 1 and shift it left 18 slots:
0000 0000 0000 0000 0000 0000 0000 0001 = 1
0000 0000 0000 0100 0000 0000 0000 0000‬ = 1 << 18

Do this for each weapon, bit-wise OR'ing the 3 together and you end up with this:
0001 0000 0100 0100 0000 0000 0000 0000

If a player is currently carrying an AWP, you can check this bit against the bitsum g_WeaponList

0000 0000 0000 0100 0000 0000 0000 0000‬ = 1 << get_user_weapon( id ) -- (assumption is he is holding AWP)
0001 0000 0100 0100 0000 0000 0000 0000 = g_WeaponList
0000 0000 0000 0100 0000 0000 0000 0000‬ = & operation between the above 2

See the matching bit in green? This would make the resulting & operation return non-zero making the condition return true if currently holding AWP.

if ( g_WeaponList & ( 1 << get_user_weapon( id ) ) )

[Napoleon_be]

Awesome, i'm starting to get a hold of it, thanks for your time & effort both of you!