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non prime numbers or composite numbers program.help plz!

This is a discussion on non prime numbers or composite numbers program.help plz! within the C Programming forums, part of the General Programming Boards category; Code: #include <stdio.h> #include <conio.h> int main() { int num; while (num<=100) /*Limit*/ {if(num%2==0) /*Condition for composite or non prime ...

  1. #1
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    non prime numbers or composite numbers program.help plz!

    Code:
    #include <stdio.h>
    #include <conio.h>
    
    int main()
    {
    int num;
    while (num<=100) /*Limit*/
    {if(num%2==0)  /*Condition for composite or non prime numbers*/
    break;
    if ((num!=2)); /* as 2 is prime so it is excluded*/
    printf("\n%d",num);/* print that non prime numbers*/
    num++; /*increment for checking next number*/
    }
    getch ();
    }

    not giving any output.what i am lacking???
    Last edited by danishzaidi; 11-15-2011 at 05:33 AM.

  2. #2
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    Not initialising num, for starters, so it's initial value is indeterminate.

    Also breaking out of the loop the first time num is divisible by 2.

    getch() and <conio.h> are also non-standard.
    Right 98% of the time, and don't care about the other 3%.

  3. #3
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    ok i defined num=1
    it just printed 1
    what should i do with the loop now? i couldn't got any idea about the breaking of loop?

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    As grumpy said before, you are breaking the loop when a number is divisible by 2, which means the program will exit the while loop (and if you start with num=1, that will occur when num=2, which is why you have only 1 in output).

    If I understand what you want to do, you want to print any non prime number, so you should just have a condition saying "if that number is not divisible by 2 and is not 2" then "print it", this should be enough.

  5. #5
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    Quote Originally Posted by Tibo-88 View Post
    As grumpy said before, you are breaking the loop when a number is divisible by 2, which means the program will exit the while loop (and if you start with num=1, that will occur when num=2, which is why you have only 1 in output).

    If I understand what you want to do, you want to print any non prime number, so you should just have a condition saying "if that number is not divisible by 2 and is not 2" then "print it", this should be enough.
    can you make changes to the code for me.

  6. #6
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    Code:
    #include <stdio.h>
    #include <conio.h>
    
    int main()
    {
    int num=1;
    while (num<=100)
    {if((num%2==0)&& (num!=2));
    printf("\n%d",num);
    num++;
    }
    getch ();
    }
    changed the code now,it is printing prime numbers also.

  7. #7
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    You should work a bit in your indentation, it will help you avoiding this kind of mistake. You have a ";" at the end of your if statement that you should remove, because now, if your condition is true, it does nothing, but the printf will be executed no matter what. Also the num<=100 is not so good, you should use num!=101 or num<101 instead.
    danishzaidi likes this.

  8. #8
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    Code:
    #include <stdio.h>
    
    
    
    p(int asal)
    {
    int y=1;
    
    
    int i4;
    for(i4=2;i4<asal;i4++)
    {
    if (asal%i4==0)
    {
    y=0;
    }
    else
    y=1*y;
    }
    return y;
    }
    
    
    
    
    int main()
    {
    int i;
    
    
    
    
    for(i=0;i<=100;i++)
    {
    if((i==0)||(i==1)) printf("%d\n",i);
    if (p(i)==1) continue;
    else printf("%d\n",i);
    
    
    }
    }

    look this. if you want to print all the prime numbers make p(i)==0
    Last edited by rac1; 11-15-2011 at 08:37 AM.

  9. #9
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    Quote Originally Posted by rac1 View Post
    Code:
    #include <stdio.h>
    
    
    
    p(int asal)
    {
    int y=1;
    
    
    int i4;
    for(i4=2;i4<asal;i4++)
    {
    if (asal%i4==0)
    {
    y=0;
    }
    else
    y=1*y;
    }
    return y;
    }
    
    
    
    
    int main()
    {
    int i;
    
    
    
    
    for(i=0;i<=100;i++)
    {
    if((i==0)||(i==1)) printf("%d\n",i);
    if (p(i)==1) continue;
    else printf("%d\n",i);
    
    
    }
    }

    look this. if you want to print all the prime numbers make p(i)==0
    I didn't run your code, but two things are bothering me while reading it :

    Code:
    //FIRST
      if(p(i)==1) continue;
      else printf("%d\n", i);
    //can be expressed as 
      if(p(i)!=1) printf("%d\n", i);
    
    //SECOND
      y=1*y; //is this really necessary?
    rac1 likes this.

  10. #10
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    Please indent and format your code correctly.


    Quote Originally Posted by rac1 View Post
    Code:
    #include <stdio.h>
    
    
    
    p(int asal)
    {
    int y=1;
    
    
    int i4;
    for(i4=2;i4<asal;i4++)
    {
    if (asal%i4==0)
    {
    y=0;
    }
    else
    y=1*y;
    }
    return y;
    }
    
    
    
    
    int main()
    {
    int i;
    
    
    
    
    for(i=0;i<=100;i++)
    {
    if((i==0)||(i==1)) printf("%d\n",i);
    if (p(i)==1) continue;
    else printf("%d\n",i);
    
    
    }
    }

    look this. if you want to print all the prime numbers make p(i)==0
    Tibo-88 likes this.

  11. #11
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    Code:
    #include <stdio.h>
    
    p(int asal)
    {
    int y=1;
    int i4;
    
    
    for(i4=2;i4<asal;i4++)
    {
       if (asal%i4==0)  y=0; 
    }
    return y;
    }
    
    int main()
    {
    int i;
    for(i=0;i<=100;i++)
    {
      if((i==0)||(i==1)) printf("%d\n",i);
    
        if (p(i)==1) continue;
        else printf("%d\n",i);
    
    }
    getchar();
    return 0;
    }
    i was using " code blocks" program, there wasnt any error.Now it is working on visual studio 2006 C program. Sorry for that reason.
    Thanks Tater, my program is very short and useful now
    Last edited by rac1; 11-15-2011 at 10:17 AM.

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