Division in C++ not working as expected

问题

I was working on something else, but everything came out as zero, so I made this minimalistic example, and the output is still 0.

#include <iostream>

int main(int argc, char** argv)
{
  double f=3/5;
  std::cout << f;
  return 0;
}

What am I missing?

回答1:

You are missing the fact that 3 and 5 are integers, so you are getting integer division. To make the compiler perform floating point division, make one of them a real number:

 double f = 3.0 / 5;

回答2:

It doesn't need to be .0, you can also do 3./5 or 3/5. or 3e+0 / 5 or 3 / 5e-0 or 0xCp-2 / 5 or... There only needs to be an indicator involved so that the compiler knows it's supposed to perform the division as floating point.

Another possibility: double f=double(3)/5. That's much more typing, but it leaves no doubt to what you are doing.

Or simply use double f=.6, that also does the trick...

回答3:

try this:

double f = 3.0/5.0;

this should fix your problem

回答4:

Try putting a .0 after one of the divisors. This will convert them into floating point literals.

回答5:

You are using integers. You can many things to make your constants double like leftaroundabout states, however that is not good clean good. It is hard to read and confusing. If you want 3 and 5 make them 3.0 and 5.0. Everyone will know what you mean if they are forced to read your code. Much of what he/she states really requires you to know C/C++ and how floats are storage to make heads or tails.

回答6:

In case, you save your generic variables with int and would like to obtain the ratio as double:

using namespace std;
int main()
{
   int x = 7;
   int y = 4;
   double ratio;
   ratio = static_cast<double>(x)/static_cast<double>(y);
   cout << "ratio =\t"<<ratio<< endl;
}

来源：https://stackoverflow.com/questions/6101084/division-in-c-not-working-as-expected

标签

c++

division