sinful condor
int i = 5, j = 2, k = 4, l = 3, m = 7, n = 8;
for ( i = 1, j = 3, k = 4; i++ < l || ++j < m && k < n; )
cout << i << " " << j << " " << k << " " << l << " " << m << endl;
OUTPUT:
2 3 4 6 7 Should j be 4 here instead as you are incrementing it by one ?
3 3 4 3 7
4 4 4 3 7
5 5 4 3 7
6 6 4 3 7
buoyant mortarBOT
restive flicker
int i = 5, j = 2, k = 4, l = 3, m = 7, n = 8;
for ( i = 1, j = 3, k = 4; i++ < l...
j doesnt increment becuase it doesnt get checked against m
in the comparison the
i++ < l is true, so there is not a need to check for the right of the or
aka
no need to check
++j < m
if left hand side is true
sinful condor
but i is = 1
restive flicker
sorry for format, im on mobile
restive flicker
but i is = 1
yeah i would be less than l
sinful condor
so you check the right side of the condition right
cause one side of the or is false
restive flicker
so you check the right side of the condition right
the right of the &&
restive flicker
cause one side of the or is false
yep
restive flicker
cause one side of the or is false
well because one side of the or is true*
sinful condor
if the left side was false how would it work
restive flicker
then it would check the right side of the ||
sinful condor
would you check the next condition or would you do the and operator and then compare it
since and has precedence over or
restive flicker
it would check ++j < m
restive flicker
im not aware that and has priority over or
i am also a newish c++ enjoyer so maybe it is true but i dont think so
sinful condor
i++ < l || (++j < m && k < n) is this technically how it is
you can put brackets around the and stuff
restive flicker
if that were the case, j would be 4 in the output. no?
its just left to right without priority
unless there are parenthesis
sinful condor
ok thanks
buoyant mortarBOT
@sinful condor Has your question been resolved? If so, run !solved :)
sinful condor
!solved