1.11M Members

C++ Division Remainder Help

 
0
 

When using C++, how do you find the REMAINDER of two numbers and store it in an integer.

I think the answer is:

X % Y = Int Z; :idea:

Please help.

Thanks :lol: ,
C++

 
0
 

like you said, "int z = x % y" gives the remainder of x / y int z. Just turn your expression around.

 
0
 

The compiler said your code was incorrect :twisted: , so I thought i would give you the orriginal question.

Here it is:

Write the definition of a function divide that takes four argumentsª and returns no valueª. The first two argumentsª are of typeª intª. The last two argumentsª argumentsª are pointers to intª and are set by the function to the quotient and remainder of dividing the first argumentª by the second argumentª. The function does not return a valueª.
The function can be used as follows:
intª numerator=42, denominator=5, quotient, remainder; divide(numerator,denominator,&quotient,&remainder); /* quotient is now 8 */ /* remainder is now 2 */

I think the answer is:

void divide(int numerator,int denominator,int* quotient, int* remainder){
*(quotient=&numerator/&denominator);
*(remainder=&denominator%&numerator);
}

But, when I run i through the compiler, these three messages appear:
In function `void divide(int, int, int *, int *)':,
invalid operands `int *' and `int *' to binary `operator /',
invalid operands `int *' and `int *' to binary `operator %'

Please help

C++

 
0
 

I think the answer is:

void divide(int numerator,int denominator,int* quotient, int* remainder){
*(quotient=&numerator/&denominator);
*(remainder=&denominator%&numerator);
}

Divide values, not memory locations:

#include <stdio.h>
  
  void divide(int numerator, int denominator, int* quotient, int* remainder)
  {
     *quotient  = numerator   / denominator;
     *remainder = denominator % numerator;
  }
  
  int main(void)
  {
     int numerator=42, denominator=5, quotient, remainder; 
     divide(numerator,denominator,&quotient,&remainder);
     printf("numerator   = %d\n", numerator);
     printf("denominator = %d\n", denominator);
     printf("quotient	= %d\n", quotient);
     printf("remainder   = %d\n", remainder);
     return 0;
  }
  
  /* my output
  numerator   = 42
  denominator = 5
  quotient	= 8
  remainder   = 5
  */

Whitespace makes code more easily readable.

 
0
 

You're correct.

Thank you SO much,
C++

 
0
 

While this post is very old, there is a mistake:

*remainder = denominator % numerator;

should be

*remainder = numerator % denominator;

http://www.cprogramming.com/tutorial/modulus.html

You
This article has been dead for over six months: Start a new discussion instead
Post:
Start New Discussion
Tags Related to this Article