C - Next 10 prime numbers - c

Question

I'm having some problems with a program I'm writing here.
It's supposed to print the next 10 prime numbers of given a value -- of course, it's not working. My logic is at fault here.
For example, if the program reads the number:
2
it's supposed to print:
3, 5, 7, 11, 13, 17, 19, 23, 29, 31
Code:
#include <stdio.h>
int main() {
int n, i, count, primenumber = 1; // primenumber is a flag
printf("Insert a number:\n");
scanf("%d", &n);
for (count = 0, n++; count < 10; n++, count++ ) {
for (i = 2; i < n; i++) {
if (n % i == 0) {
primenumber = 0;
break;
}
}
if (primenumber)
printf("%d\n", n);
}
return 0;
}
I'd be very grateful if someone were to solve this problem.
Thanks.
==UPDATE==
I did it!
The crucial change was with the primenumber flag. I inserted it inside the while, always set to 1. If set outside the while, the flag never resets and further tests won't occur -- depending on the number you give when prompted.
Here's the updated, functional code:
#include <stdio.h>
int main() {
int n, i, count = 0, primenumber;
printf("Insert a number: ");
scanf("%d", &n);
n++; // we do not want to print the prompted number
while (count < 10) {
primenumber = 1; // primenumber is set as flag
for (i = 2; i < n / 2; i++) {
if (n % i == 0) {
primenumber = 0;
break;
}
}
if (primenumber) {
printf("%d\n", n);
count++; // increment count only when prime
}
n++;
}
return 0;
}
I decided not to delete this post. I'm sure someone, someday, will find it useful.

Answer 1

Split your code a little arrange a bit more and present it
int isPrime ( int n )
{
if (n <= 1) return 0; // zero and one are not prime
unsigned int i;
for (i=2; i*i<=n; i++) {
if (n % i == 0) return 0;
}
return 1;
}
int main() {
int n, count; // primenumber is a flag
printf("Insert a number:\n");
scanf("%d", &n);
count = 0 ;
n++ ;
for ( ; count < 10; n++ ) {
if (isPrime(n)) { // if its Prime, print and increase count
printf("%d\n", n);
count ++;
}
// Check next number until we get all our numbers
}
return 0; // All Done
}

Answer 2

This is your modified working code.
In each step you should re initializing primenumber=1 otherwise it gives only one number. Also count should be increases when you find a prime otherwise not.
#include <stdio.h>
int main() {
int n, i, count, primenumber = 1; // primenumber is a flag
printf("Insert a number:\n");
scanf("%d", &n);
for (count = 0, n++; count < 10; n++ ) {
primenumber = 1;
for (i = 2; i < n; i++) {
if (n % i == 0) {
primenumber = 0;
break;
}
}
if (primenumber){
count++;
printf("%d\n", n);
}
}
return 0;
}

Answer 3

Here is a solution which will reduce the processing time, generally used for larger values of n:
#include <stdio.h>
#include <math.h>
int main() {
int n, i, count= 1, primenumber = 1, root; // primenumber is a flag
printf("Insert a number:\n");
scanf("%d", &n);
n++;
while(count!=11)
{
root= sqrt(n);//You only need to check for range 2 to square root of number. This is the key ingredient of the code to reduce time complexity
primenumber = 1;
for(i=2; i<= root; i++)
{
if(n%i==0)//As soon as it finds a number which perfectly divides it, break from loop checking n's nature
{
primenumber = 0;
break;
}
}
if(primenumber==1)
{
printf("%d \t", n);
count++;
}
n++;
}
}