C++11 std::to_string(double) - No trailing zeros

Question

Today I tried out some new functions of the C++11 STL and encountered std::to_string.

Lovely, lovely set of functions. Creating a stringstream object fo

Answer 1

std::to_string gives you no control over the format; you get the same result as sprintf with the appropriate format specifier for the type ("%f" in this case).

If you need more flexibility, then you will need a more flexible formatter - such as std::stringstream.

Answer 2

Create a custom convert function, remove the tailing zeros if necessary.

//! 2018.05.14 13:19:20 CST
#include <string>
#include <sstream>
#include <iostream>
using namespace std;

//! Create a custom convert function, remove the tailing zeros if necessary.  
template<typename T>
std::string tostring(const T &n) {
    std::ostringstream oss;
    oss << n;
    string s =  oss.str();
    int dotpos = s.find_first_of('.');
    if(dotpos!=std::string::npos){
        int ipos = s.size()-1;
        while(s[ipos]=='0' && ipos>dotpos){
            --ipos;
        }
        s.erase ( ipos + 1, std::string::npos );
    }
    return s;
}

int main(){
    std::cout<< tostring(1230)<<endl;
    std::cout<< tostring(12.30)<<endl;
}

---

The input numbers :

1230
12.30

Compile with -std=c++11, then the result:

1230
12.3

Answer 3

std::to_string(double) is defined by the standard to just return the same sequence of characters that would be generated by sprintf(buf, "%f", value). No more, no less, especially no way to tweak the format specifier. So no, there is nothing you can do.