Explinations needed

[lolikilu14]

Could someone explain to me the scope of variables and just explain functions in general I was trying to use functions as seperate sections of my program
but when I do I always get errors, any info would be helpful, thanks in
Advance.

[wildblue]

The tutorials on this site might be helpful if you haven't gotten a chance to check them out already.

http://www.cplusplus.com/doc/tutorial/functions/
http://www.cplusplus.com/doc/tutorial/namespaces/

[Disch]

just explain functions in general

A function is basically a subroutine. When you "call" a function, you program will jump to the code inside that function. When that function "returns" (or exits), the program will jump back to the code that called it.

Example:

1 2 3 4 5 6 7 8 9 10 11 12 13

void func() { cout << "inside func" << endl; } int main() { cout << "before func" << endl; func(); // <- call the function cout << "between func" << endl; func(); // <- call it again cout << "after func" << endl; }

before func inside func between func inside func after func

The program starts in your main() function. From there, each line is executed. When you call func(), the program leaves main, jumps to the func() function, and runs all the code inside that function. When it's done, the program goes right back to where it came from in main.

Could someone explain to me the scope of variables

Scope is basically the lifetime / accessibility of a variable.

For example:

void func();  // prototype the function

int main()
{
    int myvar = 5;
    func();
}

void func()
{
    cout << myvar; // <- ERROR, myvar undeclared
}

This code will generate an error.

Scope of a variable is [usually] determined by its bounding {braces}. So here... since we are declaring the variable 'myvar' inside our main function... it is only accessible inside that function. It is not 'visible' anywhere else in the program.

So func() cannot see main's 'myvar' variable because it is no longer in scope.


If you want a separate function to have access to a variable... the typical way to do that is to pass that variable as a parameter:

void func(int);  // prototype

int main()
{
    int myvar = 5;
    func( myvar ); // <- pass myvar as a parameter to func
}

void func(int myparam)
{
    cout << myparam;  // <- print the parameter
}

Here, we are creating a new variable named 'myparam' inside our 'func' function. When main calls func, it has to fill in a value for myparam. In this example... we are using 'myvar' to fill in the value.

Conceptually, it's like doing this:

int funcs_myparam = mains_myvar;

So now that we've "passed" myvar to func, we have its contents inside of the myparam variable. We can then print that through cout like normal and it'll work just fine.

[lolikilu14]

ok thanks that fixes some of the problems ive been having