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

The C++11 Standard explicitely says (21.5/7):

Returns: Each function returns a string object holding the character representation of the value of its argument that would be generated by calling sprintf(buf, fmt, val) with a format specifier of "%d", "%u", "%ld", "%lu", "%lld", "%llu", "%f", "%f", or "%Lf", respectively, where buf designates an internal character buffer of sufficient size

for the functions declared in this order:

string to_string(int val);
string to_string(unsigned val);
string to_string(long val);
string to_string(unsigned long val);
string to_string(long long val);
string to_string(unsigned long long val);
string to_string(float val);
string to_string(double val);
string to_string(long double val);

Thus, you cannot control the formatting of the resulting string.

Answer 2

With boost::to_string you cannot control the format either but it will output something closer to what you see in the screen. Same with std::lexical_cast<std::string>.

For a function-like operation with format control, use str(boost::format("...format...")% 0.33).

What's the difference between std::to_string, boost::to_string, and boost::lexical_cast<std::string>?

Answer 3

A varied solution to the problem since to_string doesn't work. "The Magic Formula" - my CS2400 teacher

std::cout.setf(ios::fixed);
std::cout.setf(ios::showpoint);
std::cout.precision(2);

const double x = 0.33, y = 42.3748;
std::cout << "$" << x << std::endl;
std::cout << "$" << y << std::endl;

Outputs:

$0.33
$42.37

any following output you do with decimal numbers will be set as so.

you can always change the setf and precision as you see fit.

Answer 4

To leave out the trailing zeros:

std::ostringstream oss;
oss << std::setprecision(8) << std::noshowpoint << double;
std::string str = oss.str();

Hope that helps.

Answer 5

If all you want to do is remove trailing zeros, well, that's easy.

std::string str = std::to_string (f);
str.erase ( str.find_last_not_of('0') + 1, std::string::npos );

Answer 6

double val
std::wstringstream wss;
wss << val;
cout << wss.str().c_str();