Question

I’m trying to write program calculating average of given numbers stored in an array. Amount of numbers should be not more than 100, and user should input them until a !int variable is given :

#include <iostream>
#include <conio.h>
using namespace std;

double average(int tab[], int i){

    int sum=0;

    for(int j=0; j<i; ++j){
            sum+=tab[j];
    }
    return (double)sum/i;

}

int main()
{
    int tab[100];
    int n=0;   
    int number=0;


    do {
       if(n < 100){
           cout << "Give " << n+1 << " number : ";
           cin >> number;
           tab[n]=number;
           number=0;
           ++n;       
       }
       else{
            break;
       }
    } while( !isdigit(number) );      

    cout << average(tab, n) << endl;

    getch();
    return 0;
}

Why after giving char, it prints me ‘Give n number:’ for all empty cells of my array ? It should end and use only given numbers.

Answer 1

isdigit tests whether the character is a digit. The test is only reached following assigning 0 to number, and 0 is a control code, not a digit, so isdigit(0) is always false, and so your while condition is always true.

 ...
       number=0;
 ...
} while( !isdigit(number) );      

Instead, test the input stream to determine whether it successfully read a value.

int main()
{
    const size_t COUNT = 100;
    int tab[COUNT];
    size_t n;   

    cin.tie(&cout); // ensures cout flushed before cin read
    // (not required if your runtime complies with that part of the standard)

    for (n = 0; n < COUNT; ++n ) {
        cout << "Give " << n+1 << " number : ";
        cin >> tab[n];

        if (!cin)
            break;
    }

    if (n > 0) // average is undefined if n == 0
        cout << average(tab, n) << endl;

    return 0;
}