I have come across a few lines of coding I do not understand, and would be grateful for clarification:
if(!(counter&7))ds->direction = ts->direction;
Thank you
views:
65answers:
3
+2
A:
The first checks if the result of a bitwise-AND on the counter with 7 is not zero, and the latter assigns the value of the direction member of one struct to the direction member of another.
Chuck
2010-10-04 15:41:50
Where "one struct" == "the struct pointed to by the pointer ds" and "another" == "the struct pointed to by the pointer ts".
sepp2k
2010-10-04 15:43:23
+3
A:
if
counteris a multiple of 8set the
directionelement of*dsequal to thedirectionelement of*ts
Paul R
2010-10-04 15:42:46
+1
A:
1) same as
if (!(counter & 7))if ((!(counter & 7)) != 0)if ((counter & 7) == 0)- if the lower 3 bits of counter are zero (or if counter is a multiple of 8)
2) same as
(*ds).direction = (*ts).direction;- set direction of
ds(must be of a struct type) to direction ofts
pmg
2010-10-04 15:44:41
In (1), I think the parenthesized part of step 4 is only correct if `counter` is unsigned...
R..
2010-10-04 15:52:14
It depends, I think, on representation. For sign and magnitude and two's complement, I think it holds (the three low bits being zero implies the number is a multiple of 8)
pmg
2010-10-04 16:06:29
Indeed. Actually I'm curious what the standard says about signed bitwise operators. I couldn't find any language specific to signed or unsigned types so I assume they're required to operate bitwise on the representations.
R..
2010-10-04 16:33:59