math - In C, why is the ratio 10.0/100 different from 0.1? -
this question has answer here:
this simple question , searched forums, couldn't find answer (i found 1 log don't think there rounding error here).
i wrote program determine value of fine range of expired products, when ratio exact, program return next fine category, ignoring = sign in conditional.
the program must return following fines:
- 0 if no product expired.
- 100 if 10% of products expired.
- 10000 if more 10% of products , 30% expired.
- 100000 if more 30% of products expired.
this code wrote:
#include <stdio.h> int calculate_fine(int ncheckedproducts, int nexpiredproducts) { int fine; float ratio; ratio=(float)nexpiredproducts/ncheckedproducts; if(nexpiredproducts==0) fine=0; else if(ratio<=0.1) fine=100; else if(ratio<=0.3) fine=10000; else fine=100000; return fine; } int main(void) { int ncheckedproducts, nexpiredproducts, fine; printf("type number of checked , expired products\n"); scanf("%d %d", &ncheckedproducts, &nexpiredproducts); fine=calculate_fine(ncheckedproducts, nexpiredproducts); printf("the fine is: %d\n", fine); return 0; } but values of 100 , 10, , 100 , 30, 10% , 30% of expired products respectively, program return wrong fine.
the teacher failed explain me why, , corrected me following function:
int calculate_fine(int ncheckedproducts, int nexpiredproducts) { int fine; if(nexpiredproducts==0) fine=0; else if(nexpiredproducts<=0.1*ncheckedproducts) fine=100; else if(nexpiredproducts<=0.3*ncheckedproducts) fine=10000; else fine=100000; return fine; } however, wish know why first 10% ratio greater 0.1, , why cannot use approach.
this is rounding issue, different might think: many finite decimal fractions not have finite binary fraction representation. thus, rounding the closest number representable floating point number of given type happens.