compute sum of two array in a array and print out the result array

Question

I have two pointer arrays for example ptr1 and ptr2, a size of 10 for each, and init random variable for every element for both. I want to compute and init two elements of ptr1 and ptr2 for each element of another array called ptr3. But when I print ptr3 out. It is not changed at all.

here is my code:

void evaluateArr(int* ptr1, int* ptr2, double* ptr3, int n) {
    // ptr3 = new double[10];
    for (int i = 0; i < n; ++i) {
        if (ptr1[i] > ptr2[i]) {
            ptr3[i] = (double)(*(ptr1 + i) * 60 / 100);
            ptr3[i] += (double)(ptr2[i] * 40 / 100);
        } else {
            ptr3[i] = (double)(*(ptr2 + i) * 60 / 100);
            ptr3[i] += (double)(ptr1[i] * 40 / 100);
        }
    }
}

void printResult(double* ptr3) {
    for (int i = 0; i < 10; ++i) {
        printf("%d ", *(ptr3 + i));
    }
}

int main() {

    int ptr1[10], ptr2[10];
    double ptr3[10];
    // init for arr1 and arr2
    for (int i = 0; i < 10; i++) {
        ptr1[i] = (rand() % 100);
        ptr2[i] = (rand() % 100);
    }
    // it prints out 0 0 0 0 0 0 0 0 0 0 
    evaluateArr(ptr1, ptr2, ptr3, 10);
    printResult(ptr3);
}

Also, all your calculations are doing integer operations. So you will not get a true double result. Change to *(ptr1 + i) * 60.0 / 100. That is, casting the final result to double is too late. Make one of the numbers double instead. — kaylum
– kaylum, Commented Apr 8, 2021 at 6:20
I would write ptr3[i] = (double)(*(ptr1 + i) * 60 / 100); ptr3[i] += (double)(ptr2[i] * 40 / 100); as ptr3[i] = (ptr1[i] * 60 + ptr2[i] * 40) / 100.0;. I think that is way more readable. — mch
– mch, Commented Apr 8, 2021 at 7:52

Rohan Kumar · Accepted Answer · 2021-04-08 08:08:29Z

1

As pointed out by people in comments there are some problems in your code:

While computing ptr3[i], since all variables in calculation expression are integer, you're getting integer as result being assigned to a double variable. You need to use a floating point number so that you get precise results:

ptr3[i] = (double)(*(ptr1 + i) * 60.0 / 100);

While printing out array in printResult(). You are printing a double value, you should use %f or %lf instead:

printf("%lf ", *(ptr3 + i));

When I apply these changes, I'm able to see ptr3 array printed properly:

c-posts : $ gcc sum2arrays.c 
c-posts : $ ./a.out 
84.599998 52.200001 69.799999 89.200001 37.400000 47.200001 77.000000 47.799999 34.000000 57.200001

edited Apr 8, 2021 at 8:08

answered Apr 8, 2021 at 6:38

Rohan Kumar

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Comments

Jonathan Leffler · Accepted Answer · 2021-04-08 14:00:00Z

0

You need to use %f or %lf to print a double value. Printing by using %d will print an integer value where your array ptr3 is of type double.

void printResult(double* ptr3) {
    for (int i = 0; i < 10; i++) {
        printf("%f ", *(ptr3 + i));
    }
}

edited Apr 8, 2021 at 14:00

Jonathan Leffler

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answered Apr 8, 2021 at 6:27

Vishwajit Rahatal

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