Comments by orbitz

Some of these pictures are really beautiful and vivid.

i guess i'm not seeing it all just yet, because i don't fully understand why the prototype is "function(int *);" i'm doing it now cuz it works.



Well think about it, a function prototype defines the...

Just a note:

the * does not really go with the datatype, but it does define its type. For instance:

int *p, q; is the same as

int *p;

int q;



to make 2 pointers it would be:

int *p, *q;



As for your...

It's like this, take a fucntion:

void blah(int r)



r is a local variable. It's value is initialized to whatever you pass it. Just because it's in a parameter list does not mean it isn't a local...

Your book will explain this concept very easily. And like I said, Thinking In C++ is free online. You aren't declaring your parameters to your function as a pointer. Try reading what I said again,...

First of all:

void main is incorrect.

It should be int main(). And because it's int main() it should have a return statment. Although in C++ that is not required.

Secondly, you should read the...