Help understanding result of Function

**deadhippo** · 04-05-2008

Hi,
I hope someone can help me understand the result of this program.

Code:

int fun3(int x);
int i = 10, j= -15, b;

main()
{
    auto int i = -20;
    int y = 7;

    y = fun3(i);

    getchar();
    return 0;

}

int fun3(int a)
{
    int b = 1;
    ++j;
    ++i;
    printf("\n (4)In fun3: Value of a=%d b=%d, i=%d and j=%d", a, b, i, j);
    return ++a;
}

It prints out "(4)In fun3: Value of a=-20 b=1, i=11 and j=-14".
I understand that because b was initialized as 1 within the function that it prints as 1 and that i and j are global variables that are incremented within the fuction and therefore have the values 11 and -14 respectively.
On the other hand I can't seem to figure out how a gets the value of -20. Can someone point me in the right direction?

**vart** · 04-05-2008

you pass -20 as a parameter (using local var i)

**deadhippo** · 04-05-2008

Originally Posted by vart

you pass -20 as a parameter (using local var i)

I'm sorry I don't know what that means.

Why is there a y in this program? It seems superfluous.
In the main function i is -20 and then y = fun3(i) so I guess that the i variable is replaced with the -20 and the int a in the external function somehow becomes -20? And therefore a = -20?

**Salem** · 04-05-2008

-Wshadow
Warn whenever a local variable shadows another local variable, parameter or global variable or whenever a built-in function is shadowed.

You've got the same variable name, in two different scopes, and you're getting confused as to which one applies at any given point in the code.

If you give everything a unique name, you'll be less confused.

**deadhippo** · 04-05-2008

Originally Posted by Salem

You've got the same variable name, in two different scopes, and you're getting confused as to which one applies at any given point in the code.

If you give everything a unique name, you'll be less confused.

It's part of a program from my textbook. I think that it is deliberately confusing to make you understand the concept but I could be wrong.

**Elysia** · 04-05-2008

A scope is defined by the brackets ({ and }). Each scope can have its own variables. Those variables are only accessible inside that scope. So variables placed outside any scope (global) are visible to the entire program.
Variables defined inside a function is only visible inside that function. And variables defined inside a specific scope is only visible inside that scope.
However, each scope can also define their own variables with the same name as variables in an outer scope. If this is the case, then the variable within the scope will hide the variable in the outer scope (that means the variable in the outer scope exists, but you can't access it because you have another variable with the same name).

**deadhippo** · 04-05-2008

Thanks for your response Elysia. I think I understand about scope and that even though it is recommended to choose different variable names this program doesn't. I think what you are saying is that when variables are declared within a particular scope global variables with the same name are ignored within that scope. Except you put it in a more elegant and easy to understand way.

I guess I am still missing something though as I really don't know how the -20 value is passed to the a.

**Elysia** · 04-05-2008

Originally Posted by deadhippo

Thanks for your response Elysia. I think I understand about scope and that even though it is recommended to choose different variable names this program doesn't. I think what you are saying is that when variables are declared within a particular scope global variables with the same name are ignored within that scope. Except you put it in a more elegant and easy to understand way.

Yes, but it happens for every scope. You can define variables within if statements with the same name as other variables and it will hide the ones in the outer scope.

I guess I am still missing something though as I really don't know how the -20 value is passed to the a.

This one is really easy. "A" is an argument to your function and you pass the variable "i" which contains the value -20, so "a" is also guaranteed to get -20 because you are passing it as an argument.
I think everyone already echoed this.

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