How to retain 0s after decimal point, after converting string to double

Question

#include <iostream>

using namespace std;

int main()
{
    string str="-1.2300";
    double d = stod(str);
    cout<<d<<"\n";
}

output: -1.23

i am expecting output as following

-1.2300

***i am expecting output as following*** That is really not how floats or doubles work. — drescherjm, Nov 08 '16 at 13:05
Then you need to do formatted output with a precision specifier. — David Hoelzer, Nov 08 '16 at 13:05

David Hoelzer · Answer 1 · 2016-11-08T13:08:38.657

0

If that is what you want, then you need to use formatted output with a precision specifier:

printf("%.4f\n", d);

or specify the precision for cout:

cout.precision(4);
cout << fixed << d << "\n";

edited Nov 08 '16 at 13:08

answered Nov 08 '16 at 13:06

David Hoelzer

Thanks, How can i do it with cout. – sks Nov 08 '16 at 13:08
Added. You will need to #include limits.h I think – David Hoelzer Nov 08 '16 at 13:08
Please lets not answer obvious duplicates. Makes cleaning them up more difficult. – NathanOliver Nov 08 '16 at 13:10
Didn't realize it was a dupe when I first answered. I rarely answer C++ questions. – David Hoelzer Nov 08 '16 at 13:16
`cout` should be `std::cout`. And `` (or more accurately the C++ equivalent, ``, has nothing to do with this. You'll need `` for `std::cout` and `` for `std::fixed`. – Andrew Nov 08 '16 at 13:35
Ah, my mistake. That's why I rarely answer C++ questions. – David Hoelzer Nov 08 '16 at 14:45

1 Answers1