for loop C work weird? - c

Question

I have a problem with for loop in C.
The purpose of the program is to find prime numbers in known quantity k.
Here is my program:
unsigned int i, k, j;
unsigned long int prime;
int _tmain(int argc, _TCHAR* argv[]) {
printf("How many prime numbers do you want to print out in order? "); scanf_s("%u", &k);
printf("%u fist prime numbers are: ", k);
i = 1;
prime = 2;
while (i <= k)
{
for (j = 2; j <= prime; j++)
{
if (prime % j == 0)
{
if (j == prime && prime != 2)
{
printf("%lu, ", prime);
i++;
}
prime++;
j = 2;
}
}
}
_getch();
return 0;
}
And when I run the program, it returns infinite sequence of numbers. But if I adds "else j++;" like this:
for (j = 2; j <= prime; j++)
{
if (prime % j == 0)
{
if (j == prime && prime != 2)
{
printf("%lu, ", prime);
i++;
}
prime++;
j = 2;
}
else j++;
}
Then the program will work out properly. I think this is kinda weird and can't explain why?
Thanks in advance (and sorry for my bad English).

Answer 1

As I mentioned in comments, the problem was caused by your having no "break" in the for-loop to cause it to return to the outer while loop when you detected a valid divisor.
This was compounded by your poor use of variable names confusing you - your original code appears to be confused about what "j" and "prime" are.
Here I've refactored your code to be cleaner and give a significant performance increase for large prime searches:
#include <stdio.h>
#define FALSE (0)
#define TRUE (!(FALSE))
int main()
{
int numPrimes = 0;
int candidate, divisor;
int isPrime;
printf("How many prime numbers do you want to print out in order? ");
scanf("%u", &numPrimes);
printf("\n%u first prime numbers are:\n", numPrimes);
if (numPrimes > 0) {
printf("2, ");
--numPrimes;
}
// we only need to test odd numbers.
for (candidate = 3; numPrimes > 0; candidate += 2) {
// N / (N/2) + 1 is always < 1, so we only have to test half the values.
// we also only need to test odd divisors
isPrime = TRUE;
for (divisor = 3; divisor <= candidate / 2; ++divisor) {
// it's not prime if testNumber divides in
if (candidate % divisor == 0) {
isPrime = FALSE;
break;
}
}
if (isPrime) {
printf("%lu, ", candidate);
--numPrimes;
}
}
printf("\n");
return 0;
}
Live demo here: http://ideone.com/yrUPBy
Which gladly tells me
How many prime numbers do you want to print out in order?
1000 first prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
37, 41, 43, 47, 53, 59, 61, 67, 71, 73,
79, 83, 89, 97, 101, 103, 107, 109, 113, 127,
...
7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841,
7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919,

Answer 2

Your loop has j <= prime as the end condition, meaning the last value of j will be equal to prime. In that case (prime % j == 0) will always be true.
When you add the extra j++ you're skipping half the values of j which accidentally skips past the problem. The program still isn't working correctly.

Answer 3

When the program prints the prime 3, the variables have the following values:
i = 1
j = 3
prime = 3
Then you increment i and prime, and set j = 2, so it's:
i = 2
j = 2
prime = 4
Now you return to the top of the loop, which does j++:
i = 2
j = 3
prime = 4
Since j <= prime, the loop continues. Since prime % j is not 0, it skips the body of that if, and returns to the top of the loop, which increments j again:
i =2
j = 4
prime = 4
This time prime % j == 0, so it prints prime, even though it's not actually a prime number.
This keeps happening for every value of prime.
When you add j++ to the else clause, it causes j to be incremented twice in the failing case. So it goes from 3 to 5, and never makes it to the case where j == prime and it prints the number. Then j <= prime fails, so the for loop terminates.