c++ - Obtaining quotient with remainder using just arithmetic operators -

i'm starting learn basics of c++ i'm stumbled upon issue.

for example, i'm trying convert celcius fahrenheit.

assuming, have fixed formula.

f = 9/5(c) + 32 

i'm aware mathematically possible break down

mathematically : (9c/5) + 32 

however, i'll not able obtain quotient reminder

so how obtain both quotient reminder using arithmetic operators or impossible

i tested out still refuses give me correct output

#include<iostream> using namespace std;  int main(){  double f; int c;  cout<<"enter degree celcius:"<<endl; cin>>c; cout<<endl;  f = 9/5 * c + 32; //incorrect ways obvious reason f = 9%5 * c + 32; //incorrect ways obvious reason  cout<<c<<"degree celcius equals to:"<<f<<endl;  } 

f = 9/5 * c + 32; //incorrect ways obvious reason 

if knew "obvious reason" why incorrect, wouldn't asking question!

it incorrect because calculation integer values. 9, 5, c, , 32 ints, , when mathematical operations ints, c++ gives result int. because integers whole numbers, not have decimal portion , cannot store remainder.

after calculation done, assign result f, double. although floating-point types have decimal portion can store remainder, you've done here convert integer result floating-point value.

you can think of code being equivalent to:

int result = 9/5 * c + 32; f = static_cast<double>(result); 

in order fix problem, need force calculation done floating-point:

f = 9.0/5.0 * static_cast<double>(c) + 32.0; 

this preserve "remainder" decimal portion of floating-point value, f.

technically, need single intermediate operation yield floating-point result. if first operation results in floating-point value, remaining operands promoted floating-point match initial operand. do:

f = 9.0/5 * c + 32; 

(9.0 double literal, 9 integer literal. adding decimal portion (.x) turns double. if had written 9.0f, float literal.)