Dividing two integers to produce a float result

Question

Possible Duplicate:
Why can't I return a double from two ints being divided

My C++ program is truncating the output of my integer devision even when I try and place the output into a float. How can I prevent this whilst keeping those to variables (a & b) as integers?

user@box:~/c/precision$ cat precision.cpp
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
  int a = 10, b = 3;
  float ans = (a/b);
  cout<<fixed<<setprecision(3);
  cout << (a/b) << endl;
  cout << ans << endl;
  return 0;
}

user@box:~/c/precision$ g++ -o precision precision.cpp 
user@box:~/c/precision$ ./precision 
3
3.000

float ans = ((float)a/b); (or static_cast to please the purists) — Vlad, Sep 16 '12 at 13:40
Yes my mistake, I did have a search around but I didn't find that post. It is a dup, my bad! — jwbensley, Sep 16 '12 at 13:42

cdiggins · Accepted Answer · 2016-08-14T18:35:30.390

158

Cast the operands to floats:

float ans = (float)a / (float)b;

edited Aug 14 '16 at 18:35

answered Sep 16 '12 at 13:41

cdiggins

17,602
7
105
102

Also, floats only have so much precision. Your integer division might require the "double precision" of a `double` – recursion.ninja Sep 16 '12 at 13:46
46

Strictly speaking you only need to cast one integer to float, althought this does make it extra clear. – Gerard May 02 '14 at 17:46
`INT K; K/3;` Here `/` will called for `INT` class, then what is the reason behind the that cast one int even nominator or denominator will work? – Asif Mushtaq Jan 16 '16 at 15:54
27

Since we are using C++ here, use `static_cast`. Note that it's possible to use the type division rule and do it simply like `float ans = static_cast(a)/b;` – Hitokage Oct 19 '17 at 06:20
1

Absolute MAD LAD! Thank you upvoted. – Wael Assaf Jan 19 '18 at 17:45
what if the integer is larger than `Float.MAX_VALUE`? – Maarten Feb 19 '20 at 08:54
@Maarten Then cast it to `double` or `long double` – Kiruahxh Oct 15 '21 at 13:10
Coming back to my comment, Float max (~2^128) > Int max (~2^31), only a lot less precise ;) – Maarten Jan 13 '22 at 21:19
how to specify the precision? e.g I would like to have only two digits after the decimal. – Mahesh Jamdade Aug 22 '22 at 01:52

Dividing two integers to produce a float result

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