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you could pass that kind of pointer with a struct, also. :)



struct array {
char *rows; //char stands for *type*
char *columns;
} Array

the function works fine, while looping.

i don't really understand why the 'if' goes all the checks, over the '1', to the hell. :)

ok, man i was only asking what's going on; don't know why your 'shrug'.

i don't understand, however, why the 'if' construct doesn't work properly. he makes his job nicely only for the input=1. on...

does it mean i have to change the structure of this code?

i'm not asking someone will do a job for me, absolutely.
i've checked the code, seems to be ok, check for the '1' is ok and the return is...

sure

it should work as in this case;

input 10.

10 is even then 10/2 = 5 (step=1)
5 is odd then 5*3+1 = 16 (step=2)
16 is even then 16/2=8 (step=3)
8 is even then 8/2=4 ...

thanks for the input :)

something strange is happening.

compile: ok

the function assigns always 1 thou, whatever is the input.

here's an example;

I think a silly syntax error then; i post the code..



#include <stdio.h>
#include <stdlib.h>

main()

{ int n;

My first post here.. :)
you should make assignments after the right declarations, it works fine for me. hope this helps.