How to check the #N bit of a s32/u32 number?

[CuteAlien]

I'm starting with 1 cause its more easy to think/say "the 7th bit is:" then "the bit in the 6th place is:"..

That's certainly up to you, but I'd recommend starting to count at 0. It's not really more difficult once you get used to it and you _will_ get used to it in the long run when programming :-)
Also when using bitoperations you often care about speed and adding the '-1' will make your operation slower.

[MasterGod]

Great, I got my answer - better then I expected

Thank you all!

[drac_gd]

Would this work ?
#define isBitSet(num,n) (( num & (0x01<<n) ) ?1:0;)
#define setBit(num,n) ( num |= (0x01<<n) )
#define clearBit(num,n) ( num &= (0x01<<n) )

shift operators are computationaly cheap

[vitek]

#define clearBit(num,n) ( num &= (0x01<<n) )

Close...

Code: Select all

#define clearBit(num,n) ( num &= ~(1<<n) )

Note that you don't need to use hexadecimal notation for 1 [0x01 is the same as 0x1, 01 and 1].

Travis

[drac_gd]

ahhh .. I see the error now.. thanks