why the prime numbers cant be displayed?

hi,i am using dev c++.
here is my code.

Code:

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#include <stdio.h>
#include <stdlib.h>
int is_prime(int n);

void main()
{
    int x;
    printf("How many prime numbers? ");
    scanf("%d", &x);
    printf("\n %d prime numbers: ",x);
   
    int j=1;
    int i=1;
    for(i=1;i<=x;i++)
    {
            printf("\n %d :",i);
            for(j=1;is_prime(j)==0;j++)
                {
                                    if(is_prime(j)==1)
                                    printf("%d",j);
                                    }
           
           
    }
   
  system("PAUSE");       
 
}
//function definition
int is_prime(int n)
{
    int k, limit;
    if(n==1)
            return 0;
    if(n==2)
            return 1;
    if(n%2==0)
              return 0;
    limit =n/2;
    for(k=3;k<=limit;k+=2)
                          if(n%k==0)
                                    return 0;
    return 1;
}

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well one thing to look at is for loop is never executed in your is_prime function

Quote:

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Originally Posted by rogster001

well one thing to look at is for loop is never executed in your is_prime function

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No, it does.
it does not print prime numbers because

Code:

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for(j=1;is_prime(j)==0;j++)

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the test condition is wrong. 0 means it is NOT prime. So it does not get executed at all.If the inner for loop is for checking whether a number is prime or not fails, you still increment 'i' thinking that it has found a prime number, when in fact it hasn't. If j is always initialized to 1 in your j for loop and a condition in your function is_prime(n)

Code:

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if(n==1)
            return 0;

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then it naturally fails every time from i to x.

All you need is:

Code:

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for(j=1; i <= x;j++)
{
                if(is_prime(j)==1) {
                                        printf("\n%d",j);
                                          i++;
                }
}

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thz, your explanation helps me much^^

BTW, you can speed up your program by changing:

Code:

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limit =n/2;

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to:

Code:

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limit = sqrt( n ) + 1;

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