CODEWITHSUNDEEP
c++arraysdynamic
Deep Kanakia
Deep Kanakia

Reputation: 11

A function is outside main,it requires use of a variable how do use it?

My function is outside main()

it's this 

void Ydisplay(int D1[])
{
 for(int i=0;i<a;i++)
  {
   cout<<"\t<<D1[i];
}

the array D1 is a dynamic array the error is 'a' is undefined it's taken from user so it has to be in main.. but is there any other option?

Upvotes: 0

Views: 81

Answers (4)

Juniar
Juniar

Reputation: 1399

I think you need to understand what you are trying to do. In this particular code, You are trying to print elements that are in the array D1. So you print the element starting from D1[0] to D1[n]. You use the for loop to traverse through each element in the array D1. int i starts at i = 0 to the last element which is i < sizeof(D1)/sizeof(int). You don`t need variable a, it make no sense with what you are trying to do. To print on each line try: cout << D1[i] << endl;

Upvotes: 0

Grijesh Chauhan
Grijesh Chauhan

Reputation: 58271

a is not know to function Ydisplay() it is local to main(), Pass value a from main.

change function syntax as:

void Ydisplay(int D1[], int a)
                         ^ add

A syntax error, missing ":

cout<<"\t" <<D1[i];
     //  ^  added

Upvotes: 0

Kerrek SB
Kerrek SB

Reputation: 477040

You have to pass the array size along as a function parameter:

void Ydisplay(std::size_t len, int D1[])
{
    for (std::size_t i = 0; i != len ;++i)
    {
        std::cout << '\t' << D1[i];
    }
}

In C++, you would use a std:vector<int>, though.

void Ydisplay(std::vector<int> const & D1)
{
    for (int n : D1)
    {
        std::cout << '\t' << n;
    }
}

Upvotes: 1

Muthu Ganapathy Nathan
Muthu Ganapathy Nathan

Reputation: 3307

Have your function in this way.,

void Ydisplay(int D1[])
{
 cin >> a; //Remove getting input from main()
 for(int i=0;i<a;i++)
  {
   cout<<'\t'<<D1[i];
}

Upvotes: 0

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