C / C++ / MFC

Purish Dwivedi wrote:

I need help

Stuart Dootson wrote:

A quick Google for "RADIUS_ATTRIBUTE_TYPE chap" leads to the relevant RFC[^], which tells us that you won't see the plain-text password in ratCHAPPassword. Rather, you'll see the encrypted password. Maybe that's what you're missing?

Stuart Dootson wrote:

Have you seen this page[^]. You need to send a challenge first, then the client should respond with a response (that'll be in the CHAP password field, I guess).

The response should be the MD5 of the CHAP ID octet, the password that the user entered and the CHAP challenge that you sent. If it is, you send back an Access-Accept, otherwise you send back an Access-Reject.

That's all documented in the RFC that I posted to you first time. Take the time and trouble to work out what your specs require you to do. I know nothing about this subject, but with some Googling, I've managed to work this out.

Stuart Dootson wrote:

how on earth could you possibly find anything like that on the internet

1. Call multiple in the loop and remove the loop in multiple:
   

   /* Program that reads a pair of numbers and determines whether the second number is multiple of the first one */
   
   #include <stdio.h>
   
   int multiple ( int j, int z); /* function prototype */
   
   int main( void ) /* function main begins program execution */
   {
      int num1, num2, x, i; /* declare variables */
   
      for ( x = 1; x <= 5; x++ )
      {	
         printf( "Enter the first number:" ); /* prompt for input */
         scanf("%d", &num1 ); /* read number from user */
   
         if ( num1 != 0)
         {
            printf( "Enter the second number:" ); /* prompt for input */
            scanf("%d", &num2 ); /* read number from user */
         } 
         else
         {
            break;
            printf( "\nBroke from loop because num1 must be greater than 0\n" ); /* break loop if num1 = = 0 */
         }
         multiple( num1, num2);
      }
   
      return 0;
   }
   
   int multiple ( int j, int z ) /* copy of the argment to function */
   {	
      int result;
   
      result = z % j;
   
      if( result == 0)
      {
         printf( "%d is multiple of %d\n", z, j );
      }
      else
      {
         printf( "%d is not multiple of %d\n", z, j );
      }
         return result;
   } /* end of multiple function */

2. Gather the numbers into two arrays and process them in one call to multiple:
   

   /* Program that reads a pair of numbers and determines whether the second number is multiple of the first one */
   
   #include <stdio.h>
   
   int multiple ( int* j, int* z, int n); /* function prototype */
   
   int main( void ) /* function main begins program execution */
   {
      int num1[5] = {0};
      int num2[5] = {0};
      int x, i; /* declare variables */
   
      for ( x = 1; x <= 5; x++ )
      {	
         printf( "Enter the first number:" ); /* prompt for input */
         scanf("%d", &(num1[x-1]) ); /* read number from user */
   
         if ( num1[x-1] != 0)
         {
            printf( "Enter the second number:" ); /* prompt for input */
            scanf("%d", &(num2[x-1]) ); /* read number from user */
         } 
         else
         {
            break;
            printf( "\nBroke from loop because num1 must be greater than 0\n" ); /* break loop if num1 = = 0 */
         }
      }
   
      multiple( num1, num2, x-1);
   
      return 0;
   }
   
   int multiple ( int* j, int* z, int numValues) /* copy of the argment to function */
   {	
      int y; /* counter */
   
      int result;
   
   
      for (y=0;y<numValues;++y)
      {
         result = z[y] % j[y];
         if( result == 0)
         {
            printf( "%d is multiple of %d\n", z[y], j[y] );
         }
         else
         {
            printf( "%d is not multiple of %d\n", z[y], j[y] );
         }
      }
      return result;
   } /* end of multiple function */

Personally, I prefer the first one - why mess with arrays when you don't have to?


Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

• you're treating num1 and num2 in main (as well as j and z in multiple) like they where array elements, but, in fact, they aren't.
• in the array-like scenario, the break on num1==0 is inappropriate, that 'exceptional' input would be better handled inside the multiple function (possibly with a continue).
• the line
  

  raeiko wrote:

  printf( "\nBroke from loop because num1 must be greater than 0\n" );

  
  is wrong: you're simply rejecting num1==0, i.e. you code accepts negative num1 values.
As already suggested by Stuart Dootson, would be probably better abandoning the array-like design: make multiply function acting on just two operands and call it inside the input loop.



If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

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