Difficulty with mod operator

Question

I would be grateful if somebody could help me with this question.

A C program contains the following declarations and initial assignments.

int i = 8,  j = 5;
float x = 0.005, y = - 0.01;
char c = 'c',  d = 'd';

Determine the value of the following expression using values assigned to the variables for the expression:

(i - 3 * j) % ( c + 2 *d)/ (x - y )

I tried this manually first:

( i- 3 * j) % ( c + 2 *d ) / ( x - y)
( 8 - 3*5) % ( 99 + 2 * 100 ) / ( 0.005 - (-0.01) )
( -7 ) % ( 299 ) / (  0.015 )

Keeping precedence and associativity in mind, I used the mod operator first:

( 292 ) / (  0.015 )

Which gave the answer 19466.66.

This does not match with the answer given in the book or when I used this in codeblocks, both of which gave the answer as - 466.6667

The codeblocks program is as below

#include <stdio.h> 
#include <stdlib.h> 
 
int main() 
{ 
    int i = 8,j = 5, c = 'c',d = 'd'; 
    float x = 0.005, y = -0.01, a = 0.0; // a is the float variable to which the value is assigned
    a = (i-3*j)%(c+2*d)/(x-y); 
    printf("%f\n",a); 
    return 0; 
}

Try not to use line numbers in your code. We'll just have to delete that, tediously, if we want to run it and test. — tadman, Mar 31 '21 at 05:38
This is an exercise in using operators and ASCII values given in the book. Thanks for the edit. — watercolour man, Mar 31 '21 at 05:48
You have an arithmetic error: `0.005 - 0.01 == -0.005`, not `-0.015`. — Nate Eldredge, Mar 31 '21 at 05:52
Also your code has `y = -0.01` but the problem has `y = 0.01`; which is it supposed to be? Finally `(-7) % 299` is not `292` but `-7`. — Nate Eldredge, Mar 31 '21 at 05:58
The problem stated y to be - 0.01. So x - y would be 0.005 - (- 0.01 ) — watercolour man, Mar 31 '21 at 06:09
Thanks for pointing out the errors . I have rectified the post. — watercolour man, Mar 31 '21 at 06:14
[Related question](https://stackoverflow.com/questions/58768191). The key takeaway from the highest-voted answer (unfortunately not the accepted answer) is that the result of `a%b` has the same sign as `a` (at least since C99). So `(-7) % (299)` is not 292. It is -7. — user3386109, Mar 31 '21 at 06:50

score 1 · Answer 1 · answered Mar 31 '21 at 05:55

The crux of this is the integer part before division:

(i-3*j)%(c+2*d)

Where that evaluates to:

(8-3*5) % (99 + 2 * 100)
(8-15) % (99 + 200)
-7 % 299

So now it depends on what definition of modulo is being used by C, as there are several it could be. C interprets this as -7 while other languages always map to the 0 to 298 range.

The rest is just simple division.

score -2 · Answer 2 · answered Mar 31 '21 at 05:39

-2

You have to use division at the end (and once) of your calculation to keep the precision.

With division, you can lose precision. See the butterfly effect Wikipedia page why it can make huge differences.

You can work with fractions and only divide when you print them.

answered Mar 31 '21 at 05:39

artit91

83
7

3

If this is butterfly effect, then the butterfly involved is [Mothra](https://en.wikipedia.org/wiki/Mothra). – tadman Mar 31 '21 at 05:58

Difficulty with mod operator

2 Answers2