Perfect numbers, arrays, verification and operation

Question

I'm trying to write a code in C which would allow a maximum input of 10 elements (natural numbers) in an array, identify all perfect numbers in the array, and do the product of all non-perfect numbers.

Euclid proved that 2^{p−1}(2^p−1) is an even perfect number whenever 2^p−1 is prime (Euclid, Prop. IX.36). For example, the first four perfect numbers are generated by the formula 2^{p−1}(2^p−1), with p a prime number, as follows: for p = 2: 2^1(2^2−1) = 6 for p = 3: 2^2(2^3−1) = 28 for p = 5: 2^4(2^5−1) = 496 for p = 7: 2^6(2^7−1) = 8128. (source: wikipedia)

When I compile the program, I get a triplicate or higher repetition of declaration of the perfect number.

e.g.:

... t[10] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 6}

'6' is a perfect number. '6' is a perfect number. '6' is a perfect number. ...

I also get a weird products.

e.g.:

... t[10] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 28}

'28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. '28' is a perfect number. ... The product of the non-perfect numbers is -1677721600

I'm really new to c, I can't seem to figure out what I'm doing wrong, but I wouldn't an handout either. Some guidance would be much appreciated.

#include <stdio.h>
#define MAX_BOUND 9  /*Array's bound*/

main() {

int i, /*array index*/
    t[i],
    d, /*divider*/ 
    sum, /*result for perfect number validation*/
    product; /*product of all non-perfect number in array*/

i = 0;

printf("Enter your natural numbers. \n");

for (i = 0; i <= MAX_BOUND; i++) {
printf("Number %d : ", i + 1);
scanf( "%d", &t[i]);
}

i = 0;
product = 1;
for (i = 0; i <= MAX_BOUND; i++) {   
    d = 1;
    sum = 0;
    while(d < t[i]) {
        if(t[i]%d == 0)
        sum = sum + d;
        d++;

            if(sum == t[i])
            printf("%d is a perfect number. \n", t[i]);
            else
            product = product * t[i];
    }
}  
printf("The product of the non-perfect numbers is %d \n", product);
getch();
}

Adam Stelmaszczyk · Answer 1 · 2012-12-02T15:09:28.623

"Weird product" (for example negative) is caused by integer overflow. Your product is int, make it bigger, long long for example.

You should use for loops with i, not while. Code checking if a number is perfect should be placed in the separate function bool isPerfect(int number).

You meant sum = 0, not somme = 0. Declaration t[i] is also wrong.

Corrected version (compile with gcc -std=c99 file.c):

#include <stdio.h>
#include <stdbool.h>
#define MAX 10

int t[MAX];

bool isPerfect(int number)
{
    int sum = 0;
    for (int d = 1; d < number; ++d) // optimization: you can iterate until sqrt(number)
    {
        if (number % d == 0)
        {
            sum += d;
        }
    }
    return sum == number;
}

int main()
{
    printf("Enter your natural numbers. \n");
    for (int i = 0; i < MAX; ++i)
    {
        printf("Number %d: ", i + 1);
        scanf("%d", &t[i]);
    }
    long long product = 1;
    for (int i = 0; i < MAX; ++i)
    {
        if (isPerfect(t[i]))
        {
            printf("%d is a perfect number. \n", t[i]);
        }
        else
        {
            product = product * t[i];
        }
    }
    printf("The product of the non-perfect numbers is %lld \n", product);
    return 0;
}

aren't **for** and **while** loops equivalent if I reinitialize **i** and add a counter (e.g. **i++**).? — , Dec 02 '12 at 14:43
long hasn't fix the problem... I guess the core of the problem are my operations (they seem alright) and my loop (way more likely). — , Dec 02 '12 at 14:49
`For`, `while`, `do-while` are logically equal (when implemented right), however when you are iterating from concrete value to concrete value, the common way to do it is with `for`. It's just one line, not 3. — Adam Stelmaszczyk, Dec 02 '12 at 14:59

Daniel Fischer · Accepted Answer · 2012-12-02T15:05:11.377

In your declaration of the array, you have undefined behaviour because you used the wrong size:

main() {

    int i, 
        t[i],
        d, /*divider*/ 
        sum, 
        product;

    i = 0;

    printf("Enter your natural numbers. \n");

    while (i <= 9) {
        printf("Number %d : ", i + 1);
        scanf( "%d", &t[i]);
        i++;
    }

You probably meant to declare

t[MAX_BOUND+1];

(MAX_BOUND would be wrong, since you use the element t[MAX_BOUND]).

At the point where t is declared, i has indeterminate value (not unlikely to be 0).

With indeterminate array size, the accesses t[i] produce yet more indeterminate values (and again are undefined behaviour if i >= sizeof t / sizeof t[0]).

The printing part,

        if(sum == t[i])
        printf("%d is a perfect number. \n", t[i]);
        else
        product = product * t[i];

should be moved after the loop used to determine the divisor sum. With that inside the loop, you multiply product with t[i] a total of t[i] - 1 times (or t[i] - 2 if one of the intermediate sums equals t[i]) if t[i] is not perfect, and t[i]/2-1 times if t[i] is perfect. Also, you print t[i]/2 times for perfect numbers, and once for abundant numbers if one of the intermediate sums equals t[i] (I'm ignoring the theoretical possibility of odd perfect numbers, if there are any, they are far too large for int).

Doing that produces correct output here.

So my mistake was my loop integration. Thank you, you were right! — , Dec 02 '12 at 15:10

Perfect numbers, arrays, verification and operation

2 Answers2