CODEWITHSUNDEEP
c++arraysmultidimensional-array
diggers3
diggers3

Reputation: 229

2D array of Doubles returning 0

I have this 2D array and I am checking the value at [5][5]. The value should be 1/11 or 0.0909 as a double. The issue is that I am getting 0 as a return value.

I am checking to make sure I have implemented the 2D array correctly. This is within main.

    //start clock
clock_t time = clock();

int n;

printf("Please enter a value for n: ");

//get n
cin >> n;

printf("\nn = %i \n", n);

//allocate space for matrix A
double **A_n = new double*[n];
for (int k = 0; k < n; ++k) {
    A_n[k] = new double[n];
}

for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
        A_n[i-1][j-1] = (double)(1/(i+j-1));
    }
}

//Test Matrix A_n
printf("A_n[5][5] = %e \n", A_n[5][5]);

Upvotes: 1

Views: 107

Answers (2)

North-Pole
North-Pole

Reputation: 630

Here is your problem:

    (double)(1/(i+j-1))

You divide an int with an int which will produce an int and after that you cast it to double. At the point of casting it is already a 0. One of the sides of the expression has to be a double. You could do that

    1 / (double)( i + j + 1 )

or

    1.0 / ( i + j + 1 )

Upvotes: 1

Carl Norum
Carl Norum

Reputation: 224944

Your typecast is in the wrong place. Your expression

(double)(1/(i+j-1))

will almost always be 0 due to integer division. You will get 1 out when i == j == 1. You probably want:

1.0/(i+j-1)

Upvotes: 4

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