在C ++中将Double转换为字符串的问题

Question

So I know setprecision(int n) should be used when printing a double value with precision n . However, I've run into a problem on a project that I'm working on that is akin to this code:

#include <iostream>
#include <iomanip>

using namespace std;

int main() {
    double var = 1.0000001;
    cout << setprecision(10) << var << endl;
    string str = to_string(var);
    cout << str << endl;
    return 0;
}

Here is the output:

1.0000001
1.000000

In the project I'm working on, I need to save the double value as a string, and it will occasionally need more than six decimal places of precision. Here, precision is clearly lost in the conversion. Any pointers would be greatly appreciated.

Answer 1

Here's a reference for to_string . Note that it produces "As many digits are written as needed to represent the integral part, followed by the decimal-point character and six decimal digits".

The rest of the question is a dupe of lots of SO ansers - see here , for example.

Answer 2

我相信您正在寻找的是std :: fixed，如下所述： http : //www.cplusplus.com/reference/ios/fixed/

Answer 3

Using ostringstream will solve your problem.

#include <sstream>

double var = 1.0000001;
ostringstream os;
os << setprecision(10) << var;
string str = os.str();
cout << str << endl; // this should print 1.0000001