cannot convert parameter 1 from 'int *_w64' to 'int'

[mng2nf]

i'm having problems with this portion of code... the entire program is an invoice for a carpet installation asking for the room dimensions and various rates and amounts to calculate how much to charge someone for service. the assignment requires you to pass at least one parameter by reference, which would be in this function. i've tried a ton of different ways to do it but clearly i don't understand the whole function scene because i just can't get it to work and my professor insists it's how i'm passing the parameter. the error with this code says "error c2664: cannot convert parameter 1 from 'int *_w64' to 'int'.

//declaring functions
int GetLength();
int GetWidth();
int CalculateArea(int Length, int Width);
 
 
int main()
{
    int Length, Width, Area;
 
//print functions
    Length=GetLength();
    Width=GetWidth();
    Area=CalculateArea(&Length, &Width);
 
    return 0;
}
 
//ask user for length, return as int
int GetLength(int &Length)
{
 
        cout << "What is the length of the room?" ;
        cin >> Length;
 
        while (Length <0)
        { cout << "Please enter a positive number." << endl;
        cin >> Length;
        }
 
    return Length;
}
 
//ask user for width, return as int
int GetWidth(int &Width)
{
 
        cout << "What is the width of the room?" ;
        cin >> Width;
 
        while (Width <0)
        { cout << "Please enter a positive number." << endl;
        cin >> Width;
        }
 
 
    return Width;
}
 
int CalculateArea(int Length, int Width)
{
        int Area;
        Area=Length*Width;
 
        return (Area);
 
 
}

[Ganon11]

When you declare your functions, you have them returning an int and accepting no parameters. Look down at your function definitions - they say they are returning an int and accepting an integer by reference. See a problem here?

You don't need the return value if you're going to use pass-by-reference and get the values in the functions. But if you change the program to work this way, you'll need to modify your main() function.

[Laharl]

Besides that, when you use the & operator as a prefix operator, like &width, you're actually passing a pointer, because & is the address-of operator. To fix that, when you call the function, remove the &.

[mng2nf]

Besides that, when you use the & operator as a prefix operator, like &width, you're actually passing a pointer, because & is the address-of operator. To fix that, when you call the function, remove the &.

i'm supposed to pass it like that... that's the one my teacher wants passed using the address instead of the variable

[pbmods]

Changed thread title to better describe the problem (did you know that threads whose titles do not follow the Posting Guidelines actually get FEWER responses?).

[Laharl]

i'm supposed to pass it like that... that's the one my teacher wants passed using the address instead of the variable

With that syntax, you're passing by pointer, not by reference, in which case, in the function definition, change the & to *, so that you are passing pointers rather than &. You will need to dereference the pointers before you can change anything, however.

[Ganon11]

Ermm....If he has something like

void getWidth(int& Width)

as a function header, that will work just fine, Laharl. C++ knows that he's passing a pointer, but does all the work behind the scenes to do it, so that he just has to use:

getWidth(width)

in main. C++ equates this to passing &width, and in the function, dereferences that pointer immediately, so you can treat it like a normal variable.

So there's no need to change the & to a *.

[weaknessforcats]

but clearly i don't understand the whole function scene

It's quite simple.

Image that you are the function. Your arguments come in the mailbox in the form of letters. One of the letters has a questionaire for you to fill out.

It should be obvious that this is a copy of the original questionaire.

When you return it, you make a copy, put it in an envelope, and mail it.

The questionaire that you received goes into your wastebasket.

This is the simple call by value that is used for variables.

I may happen that you do not receive a questionaire but instead your mail has an envelope with a URL in it. You use the URL to locate the the questionaire an complete it online.

Here the URL is a copy of the URL the sender has. But the URL is not the3 questionaire, rather, it is the address of the questionaire. By using the address you can locate and use the original questionaire.

This is the simple call by value that is used for pointers to variables.

Your original function was:

int CalculateArea(int Length, int Width);

So, you call the function with the variables themselves. A copy is made and the copy shows up in the mail for CalculateArea.

However, your call was:

Area=CalculateArea(&Length, &Width);

and you passed in the addresses of the variables instead of the variables themselves. To do this, your function needs to be set up to receive pointers:

int CalculateArea(int* Length, int* Width);