Cout doubles without cutting variables

Question

Here is some "pseudo" c++ code

double var = 5.5;
cout << var << endl;
double var2 = 6.0;
cout << var2 << endl;

5.5
6

The problem is that some code might expect that it's 6.0. Does someone has an idea how to change the output stream.

Okay here is some example code i tried

 double t = 6.0;
  while (t > 0.1) {
    t = t - 0.1;
    cout << setprecision(2) << t << endl;
  }

The output is

5.9
5.8
...
5.2
5.1
5
4.9
4.8
4.7

score 4 · Accepted Answer · edited May 23 '17 at 10:08

4

Use std::setprecision along with std::fixed:

std::cout << std::fixed << std::setprecision(1) << var2 << std::endl;

See demo : http://ideone.com/Arz85

Only std::setprecision(1) would not work. You've to use std::fixed as well.

Don't forget to include this:

#include <iomanip>

I guess you've written using namespace std in your code. If so, then don't do that. Write fully qualified names, such as std::cout, instead of cout. See these topics:

edited May 23 '17 at 10:08

Community

1
1

answered Dec 12 '11 at 12:50

Nawaz

353,942
115
666
851

Thanks this made. Especially the fixed() one was important! – Daniel Wehner Dec 12 '11 at 13:01
I agree that using namespace std is bad here, but it really helps to be able to read the code, at least from my perspective. – Daniel Wehner Dec 12 '11 at 13:02

score 0 · Answer 2 · answered Dec 12 '11 at 12:50

0

Use ios_base flags, take a look at the C++ reference.

answered Dec 12 '11 at 12:50

Sebastian

8,046
2
34
58

Cout doubles without cutting variables

2 Answers2