Converting Program to return a Pointer with a structure

Question

Can someone explain what i need to do in order to convert this program to return a pointer in the enterCubScouts function, I've tried everything i know of and nothing is working. I Read something about useing -> instead of the normal * but I'm kind of confused. Do you use -> in tandem with * or just ->.

 #include <iostream>
using namespace std;

struct CubScouts
{
    string name;
    int schoolGrade;
    string denName;
};

CubScouts *enterCubScouts(CubScouts *scouts[], int);
void printCubScouts(CubScouts scouts[], int);
int main()
{

    int numScouts;
    cout << "\n\nHow many cub scouts are in your pack?\n";
    cin >> numScouts;
    cin.ignore();
    CubScouts scouts[numScouts];

    enterCubScouts(scouts, numScouts);


    printCubScouts(scouts, numScouts);

    return 0;
}

CubScouts *enterCubScouts(CubScouts *scouts[],int size)
{
    for(int x=0; x<size; x++)
    {
            cout << "\nCUB SCOUT " << x+1 << ": \n";
            cout << "NAME: ";
            getline(cin, scouts[x].name);

            cout << "\n\nGRADE (1-5): ";
            cin >> scouts[x].schoolGrade;

            cout << "\n\nDEN NAME: ";
            cin.ignore();
            getline(cin, scouts[x].denName);
            cin.sync();
    }   
    return *scouts; // This needs to be a pointer 
}

For starters, enterCubScouts should return a pointer to CubScouts, not the struct; so you want CubScouts *enterCubScouts(blah, blah) — Jiminion
– Jiminion, Commented Apr 30, 2014 at 1:45
You're "new" call doesn't make sense. Maybe you should name your struct? Or else use struct when referencing it again. — Jiminion
– Jiminion, Commented Apr 30, 2014 at 1:53
@Jim ive deleted that in my code already, that was one of my failed attempts to figure this out — user3587476
– user3587476, Commented Apr 30, 2014 at 2:00
What are you returning the pointer for? Is the loop to be in the main call rather than the enterCubScouts() call? — Jiminion
– Jiminion, Commented Apr 30, 2014 at 2:02
"which will accept the pointer to the allocated array and the number of cub scouts." the extent of what i know about adding this part into the program, it worked fine without the pointer but im required to have it. — user3587476
– user3587476, Commented Apr 30, 2014 at 2:06

user1274820 · Accepted Answer · 2014-04-30 02:08:11Z

1

http://www.functionx.com/cpp/examples/returnpointer.htm

CubScouts * enterCubScouts(CubScouts *scouts[], int size)

Maybe?

I think you just need to add an asterisk

Going to try this in a c++ compiler; can't believe I don't have one on this machine

This worked for me:

CubScouts * enterCubScouts(CubScouts scouts[], int size)
{
    for (int x = 0; x<size; x++)
    {
        cout << "\nCUB SCOUT " << x + 1 << ": \n";
        cout << "NAME: ";
        getline(cin, scouts[x].name);

        cout << "\n\nGRADE (1-5): ";
        cin >> scouts[x].schoolGrade;

        cout << "\n\nDEN NAME: ";
        cin.ignore();

        cin.sync();
    }
    return scouts; // This needs to be a pointer 
}

void printCubScouts(CubScouts scouts[], int size)
{
    for (int x = 0; x<size; x++)
    {
         cout << scouts[x].name << " " << scouts[x].denName << " " << scouts[x].schoolGrade;
    }
}

not sure if it's doing what you want it to

edited Apr 30, 2014 at 2:08

answered Apr 30, 2014 at 1:44

user1274820

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5 Comments

user1274820 Over a year ago

return &scouts; ? sorry I am trying to get it to compile but I haven't used c++ in so long

user3587476 Over a year ago

no dice on &scouts and heres a virtual compiler if you dont want to install one, onlinecompiler.net

user1274820 Over a year ago

return scouts; ? thanks for the virtual compiler, I have it in visual studio now but it doesn't like getline and I'm trying to figure out why v.v I'll try your link

user3587476 Over a year ago

Leaves this error 13.cpp: In function 'int main()': 13.cpp:41:34: error: cannot convert 'CubScouts*' to 'CubScouts**' for argumen t '1' to 'CubScouts* enterCubScouts(CubScouts**, int)'

user1274820 Over a year ago

Take the asterisk off of scouts: CubScouts * enterCubScouts(CubScouts scouts[], int size)

M.M · Accepted Answer · 2014-04-30 02:07:01Z

You don't even use the returned value so you might as well make it void.

This line:

CubScouts *enterCubScouts(CubScouts *scouts[],int size)

should be:

CubScouts *enterCubScouts(CubScouts scouts[],int size)

(and update the prototype, too). The way you have it currently, should generate a compiler error. Also , this line:

CubScouts scouts[numScouts];

should generate a compiler error, because arrays cannot have run-time size in C++. Try invoking your compiler in standard mode. If it seems to work for you, I'd still advise using standard containers instead of your compiler extension, as we do have a document that describes exactly how standard containers have to work.

Anyway... logically it doesn't make a lot of sense to return a pointer to a scout; since your function has no code to detect input failure. What would main() do with this pointer?

If you added some error-checking then you could return a pointer indicating the end of the list of scouts that were entered successfully, e.g.

return scouts + size;

Jiminion · Accepted Answer · 2014-04-30 02:18:21Z

This compiles at least -- not sure what the print is/ supposed to do:

#include <iostream>
using namespace std;

struct CubScouts
{
string name;
int schoolGrade;
string denName;
};

CubScouts *enterCubScouts(int x);
//void printCubScouts(CubScouts*[], int);

int main()
{

int numScouts;
cout << "nnHow many cub scouts are in your pack?n";
cin >> numScouts;
cin.ignore();
CubScouts *scouts[numScouts];

for(int x=0; x<numScouts; x++){


scouts[x] = enterCubScouts(x);

}

//printCubScouts(scouts, numScouts);

return 0;
}

CubScouts *enterCubScouts(int x)
{
 CubScouts *scout = new CubScouts;
        cout << "nCUB SCOUT " << x+1 << ": n";
        cout << "NAME: ";
        getline(cin, scout->name);

        cout << "nnGRADE (1-5): ";
        cin >> scout->schoolGrade;

        cout << "nnDEN NAME: ";
        cin.ignore();
        getline(cin, scout->denName);
        cin.sync();

return scout; // This needs to be a pointer 
}

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