C programming flow in relation to Stack

**mousanony** · 08-04-2010

Hi, I was reading "Let us C" by Yashwant Kanektar (Thanks for this great book from the recomended books thread!) when i came accross this problem:

predict the outcome:

Code:

#include <stdio.h>

int i = 0;

int main()
{
printf("\nmain's i = %d", i);
i++;
val();
printf("\nmain's i = %d", i);
val();

}

val()
{
i = 100;
printf("\nval's i = %d", i);
i++;
}

When i compiled the program and run it, this was what came out:

main's i = 0
val's i = 100
main's = 101
val's i = 100

I thought the answer should have been
main's i = 0
val's i = 101
main's = 101
val's i = 100

i dont get the flow of the program. I mean, I understand that whenever val() is called, i resets to 100, however, I don't get why main becomes 101.

Plus, the book talks about "stack" which is responsible for the flow of control. Can someone explain to me what happens inside the computer (the flow of the program when the stacking system is implemented).

BTW: I have already read up on the internet about stack, I know that stack is LIFO (last in first out), sadly, I think the information I have obtained is still lacking.
In addition, I have already read about the different variable storage types.

I don’t get why the answer came at like that in the program.
Can anyone please explain this to me? Thanks

**laserlight** · 08-04-2010

Originally Posted by mousanony

I mean, I understand that whenever val() is called, i resets to 100, however, I don't get why main becomes 101.

Observe the i++; statement in val().

By the way, you should indent the code. The val function should also be forward declared with a void return type, and you should specify void as the parameter list to indicate that it takes no arguments rather than an unknown number of arguments, although that is not a problem here:

Code:

#include <stdio.h>

void val(void);

int i = 0;

int main(void)
{
    printf("\nmain's i = %d", i);
    i++;
    val();
    printf("\nmain's i = %d", i);
    val();

    return 0; /* optional in C99 */
}

void val(void)
{
    i = 100;
    printf("\nval's i = %d", i);
    i++;
}

That said, maybe this can be a lesson to you as to why you should avoid global variables: they can make it more difficult for you to reason about your program.

**ComputerWarrior** · 08-04-2010

well, the outcome is simple to predict. 'i' is set to zero
at the start of your program, then in main() 'i' is printed
and then incremented by one. main() calls val() then,
val() sets 'i' to 100, prints 'i', then increments 'i' by one.
in main, it prints 'i' again. so the algorithm would be:

Code:

i = 0

int main()
{
print 'i'
increment i
val()
print 'i'
val()
return 0
}

val()
{ 
 i = 100
 print 'i'
 i++
}

**Elysia** · 08-04-2010

You are taking the code and making it worse.
- Code is not indented.
- main and val lacks a return type.
- val is not forward declared.
If you're going to show code, at least set a good example.

**ComputerWarrior** · 08-04-2010

Elysia, im showing the code 'as is'. im pretty sure you saw i said 'algorithm' in
my post above.

**rags_to_riches** · 08-04-2010

"Let us C" by Yashwant Kanektar (Thanks for this great book from the recomended books thread!)

Oy vey...I can't believe someone recommended that piece of Turbo C-pushing crap.

**Elysia** · 08-04-2010

You deleted the return type of main, though I see it's there now.
And it's not indented. Indentation is required to make things readable. Even if it's just an algorithm or example, try indenting.

Shall we have a mod delete that recommendation?

**laserlight** · 08-04-2010

im showing the code 'as is'. im pretty sure you saw i said 'algorithm' in
my post above.

Yeah, except that in this case presenting pseudocode does not really make things clearer since the code itself already exists and is simple enough to analyse by inspection.

Anyway, I shall now call everyone to support "pseudo-Python" for pseudocode: you automatically are inclined to indent properly

**mousanony** · 08-05-2010

Thanks everyone, I really appreciate people who take the time to help others and explain stuff thoroughly.
BTW , laserlight, when you say forward declare the function, did you mean to give it a function prototype? Do they mean the same thing?

**laserlight** · 08-05-2010

Originally Posted by mousanony

BTW , laserlight, when you say forward declare the function, did you mean to give it a function prototype? Do they mean the same thing?

Yes. There may or may not be minor technical differences between the two terms, but here I mean them to be the same thing.

**mousanony** · 08-06-2010

Ok, got it. Thanks

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