difference between if(u>0) and if((u&1)>0)

[Alfred Kemety]

if us is a long value that is always positive (unsigned long in C realm)
What is the difference between if(u>0) and if((u&1)>0)
I mean (u&1) will only return 0 if "u" actually equals 0 if it is equal to 1 or 2 (10) or 3 (11) or 4 (100) then it will not return 0. are am I mistaken?
so what's the difference? There's a code snippet in a book that uses C to express algorithms and it uses if(u&1) which is equivilant to - I guess - if((u&1)>0) in java...
what am I missing?

[Barry Gaunt]

Hi Alfred, welcome back!
u & 1 is telling you the state of the low-order bit of the (unsigned or signed) longword. u & 1 will be 0 or 1 depending on whether the low-order bit of u is 0 or 1 respectively.
In C: if( x ){ // here if x != 0 } else { // here if x == 0 }
Is that what you are asking?
-Barry

[Alfred Kemety]

Ah, so it is some kind of check whether u is odd or even, true if odd and false if even...
Thanks a lot.. I just got confused with the code for a second.. I guess I should take a break, been reading for the last 5 hours.. simple stuff seems really puzzely when one is tired...

[Barry Gaunt]

Originally posted by Alfred Kemety:
Ah, so it is some kind of check whether u is odd or even, true if odd and false if even...

Yes, in a way, but odd and even are too, what's the word, specific. It could have been u & 2, and in that case it is checking the state of 2nd low order bit. If that u was some kind of device status register then the bit could represent whether or not interrupts were enabled.