Can someone explain to me why the out put of this is '001012012301234'? My professor didn't do a very good job -_-Code:#include <stdio.h> void a(int num); void b(int num); int main() { a(6); return 0; } void a(int num) { int index; for (index=0; index<num; index++) b(index); } void b(int num) { int index=0; for (index=0; index<num; index++) printf("%d", index); }
Isn't it obvious? For integer N > 0:
a(N) prints: 001:012:0123:01234:...:012...(N-3)(N-2)
Devoted my life to programming...
It's a glorified nested loop, and adding newlines helps see where the breaks are:
For each iteration of the outer loop, you print all numbers from 0 to the current index of the outer loop:Code:int index, index2; for (index = 0; index < num; index++) { for (index2 = 0; index2 < index; index2++) printf("%d", index2); printf("\n"); }
>Isn't it obvious?Code:0 01 012 0123 01234 ...
As amazing as it may seem to you, most things are not obvious when one is a beginner.
My best code is written with the delete key.
Sorry, wrong answer. You ask why. Well it is because the bigger integer passed to b() is N-1 and the bigger integer printed by b() is N-2.
Devoted my life to programming...
I know, i've been a beginner before... Sorry, it just came so naturally to me that i forgot how confusing things can be when a newbie.Originally Posted by Prelude
Devoted my life to programming...
That code is very confusing. Maybe you should rename your variables. You have two different variables named index and num. What's up with that?
Your best bet is to become a code tracer... follow the values of the variables through each iteration of the loops in each function... you'll figure it out.
example:
pass 1... a (6) -> a num = 6 -> a index = 0 -> b index = 0 -> print 0
pass 2... a index = 0 -> b num = 0 -> b index = 1 -> return
pass 3 ... a index = 1 -> b num = 1 -> b index = 0 -> print 0
pass4 ... a index = 1 -> b num = 1 -> b index = 1 -> print 1
and so on...
