Cout long double issue

Question

So, I\'m working on a C++ project. I have a var of long double type and assigned it a value like \"1.02\"

Then, I try to use cout to print it and the result is: -0

Answer 1

I see nothing wrong at all in the code. I just put it into a standard format and it works. Here is the code assuming what you posted is the entire thing.

#include 

using namespace std;

int main(){
    long double var = 1.0202;
    cout.precision(5);
    cout << var << endl;
}

I hope this answers your question.

Edit: P.S. Shorter, the better, so I have a better solution (opinionated).

#include 
#include 
using namespace std;

int main(){
    long double var = 1.0202;
    //cout.precision(5);
    cout << setprecision(5) << var << endl;
}

I think this one is better since it is shorter. I would also recommend using printf if you are doing more complex decimal stuff since printf can choose which variables (if you have multiple) have decimals or how much.