How to get a void function to work with another void function in C++?

Question

I put '(RE)' in void F() , but I approached one funny problem with my program. When I put (RE) in my void F() , my void RE() is coded above the void F() , how would the void RE() be able to know the void F() ? Visual Studio wont let me run my program this way. I thought they were declared as void functions outside main() so I assumed those would work anywhere.

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void F()
{
    if (nextChar() == 'a')
        match('a');
    else if (nextChar() == 'b')
        match('b');
    else if (nextChar() == 'c')
        match('c');
    else if (nextChar() == 'd')
        match('d');
    else if (nextChar() == 'a')
    {
        match('(');
        RE();      //HOW????
        match(')');
    }
}

void RE()
{
    if (nextChar() == 'a')
    {
        RE();
        RE();
    }
    else if (nextChar() == 'a')
    {
        RE();
        match('|');
        RE();
    }
    else if (nextChar() == 'a')
    {
        RE();
        match('*');
    }
    else if (nextChar() == 'a')
        F();                   //How????
}


int main()

Answer 1

Functions can have declarations and definitions . To be able to call a function, all you code needs is to be able to see a declaration.

So, provide declarations for both RE and F and then define them.

void RE();
void F();

//RE and F definitions here.

Answer 2

Put a declaration of RE() before F() :

void RE();

void F()
{
    ...
}

void RE() 
{
    ...
}