c program that inputs 5 pairs of integers and determine if the second one is multiple of the first one

raeiko

Hello everybody, i need some help with my code. The program seems to work,the problem is that it only outputs the result for the last pair of numbers and repeat it for 5 times. Could you please help me to understand what i'm doing wrong? Thanks a lot, raeiko /* 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_s("%d", &num1 ); /* read number from user */ if ( num1 != 0){ printf( "Enter the second number:" ); /* prompt for input */ scanf_s("%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 y; /* counter */ int result; result = z % j; for ( y = 1; y <= 5; y++ ) 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 */

Stuart Dootson

You need to put the multiple( num1, num2); call in the for loop - it's not in the loop in the code you've posted.

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raeiko

If i don't ask too much,could you please show me? This thing has been driving me crazy for one week...and i guess this is still nothing in comparison to advanced C...I feel a bit stupid... Thanks again, raeiko

CPallini

IMHO your code has several flaws.

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. :)

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Stuart Dootson

I see what you've done. You've got the loop in main and multiple, but you're passing single integers into multiple. There are two solutions (both built & tested under darwin-gcc-4.0.1):

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 */
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 multip

raeiko

Thank you so much!!! Now i see what was wrong:-) raeiko

David Crow

Check out your multiple() function. Notice that it is called outside of the loop, so it is only going to work on the last two values of num1 and num2. Also, why does the multiple() function have a for loop?

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