CODEWITHSUNDEEP
c++arrays
user3739376
user3739376

Reputation: 1

Why doesn't my array store any input values?

I want to input any number into array b[] by number of numCase times. 

#include <iostream>
using namespace std;

//entry point

int main()

{


//Declarations

int b[20]; // array size 20 ( limit of inputs)
int c = 0;
int numCase;
int input;





 cout << "ENTER NUMBER OF CASES (MAXIMUM NUMBER OF 20):  \n";
 cin >> numCase;


//checks that numCase is less than or equal to (20) and does not exceed
if (numCase < 21)
        {



// gets input number based on the numCase

do
{


cout << "ENTER A NUMBER (MAXIMUM OF 5 DIGITS): \n";
                cin >> input;
                cout << "\n";
                b[c] = input;
                c++;



} while (c != numCase);


cout << b[c] ;  //  this is my problem it OUTPUTS RANDOM VALUE, 
     //but i can  see on my watch list that b has the values of my input.
}

}

Upvotes: 0

Views: 95

Answers (5)

Shubham
Shubham

Reputation: 165

If you want to display all the numbers stored in array b[] then you may write your code as

for(int i=0;i<=20;i++)
{ if(b[i]<101)   //This will exclude all the values which are greater than 101
   {cout<<"\n"<<b[i];}}

Upvotes: 0

KennethLJJ
KennethLJJ

Reputation: 157

I think this is what you might be looking for:

for (int i=0; i<numCase; i++)
{
    if(b[i] >= x) //x is a variable that u can set as a limit. eg. 700
    {
        cout<<"\n"<<b[i];
    }
}

Hope it helps

Upvotes: 0

rajenpandit
rajenpandit

Reputation: 1361

cout << b[c] ; // in this statement c already reached to numCase where you have not assigned any value

Upvotes: 0

RichieHindle
RichieHindle

Reputation: 281855

You're filling entries 0 toN of b, and then printing entry N+1, which you haven't filled in.

Upvotes: 2

suspectus
suspectus

Reputation: 17288

The variable c should be initialised back to zero.

} while (c != numCase);

c = 0;

Upvotes: 1

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