CODEWITHSUNDEEP
c++structpass-by-reference
Gary Nunchucks
Gary Nunchucks

Reputation: 3

Passing a struct member to function C++

Ok,so I have a struct and I need to create a function that increases the amount of a book.After calling the function for each book ,I will call printBooks which I can do just fine.I know it's a very simple program but I just couldn't do it so any help is appreciated

#include <iostream>
using namespace std;

#define BOOKS 3

struct Book
{
    string title;
    string isbn;
    int amount;
} books [BOOKS];

void printBooks(Book b);
void addAmount(Book &book,int amount);


int main()
{
    int i;

    for(int i = 0;i < BOOKS; i++)
    {
        cout << "Enter isbn : ";
        cin >> books[i].isbn;

        cout << "Enter title : ";
        cin >> books[i].title;

        cout << "Enter amount : ";
        cin >> books[i].amount;

    }


    cout << "\nThe number of books after adding amount by one :\n";

    for(i = 0;i < BOOKS; i++)
    {
        addAmount(); // intentionally left blank.don't know what to put
        printBooks(books[i]);
    }

    return 0;
}

void printBooks(Book b)
{
    cout << b.isbn << endl;
    cout << b.title << endl;
    cout << b.amount << endl;
}
void addAmount(Book &book,int amount)
{
    book.amount++;
}

Upvotes: 0

Views: 4344

Answers (2)

Maksim Skurydzin
Maksim Skurydzin

Reputation: 10541

void addAmount(Book &book)
{
    book.amount++;
}

and then

addAmount(books[i]);

Btw, you don't really need to pass a structure by value to printBooks. Just use a const reference instead. The calling code doesn't need to be modified.

void printBooks(const Book &b)
{
    cout << b.isbn << endl;
    cout << b.title << endl;
    cout << b.amount << endl;
}

Upvotes: 1

Luchian Grigore
Luchian Grigore

Reputation: 258548

You're calling addAmount(); without parameters. You probably mean

addAmount(books[i], 42);

Also consider changing the printBooks signature to

void printBook(const Book& b)

or, even better, making it a member function.

Upvotes: 6

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