How do I convert a double into a string in C++?

问题:

I need to store a double as a string. I know I can use printf if I wanted to display it, but I just want to store it in a string variable so that I can store it in a map later (as the value, not the key).

回答1:

The boost (tm) way:

std::string str = boost::lexical_cast<std::string>(dbl);

The Standard C++ way:

std::ostringstream strs;
strs << dbl;
std::string str = strs.str();

Note: Don\'t forget #include <sstream>

回答2:

// The C way:
char buffer[32];
snprintf(buffer, sizeof(buffer), \"%g\", myDoubleVar);

// The C++03 way:
std::ostringstream sstream;
sstream << myDoubleVar;
std::string varAsString = sstream.str();

// The C++11 way:
std::string varAsString = std::to_string(myDoubleVar);

// The boost way:
std::string varAsString = boost::lexical_cast<std::string>(myDoubleVar);

回答3:

The Standard C++11 way (if you don\'t care about the output format):

#include <string>

auto str = std::to_string(42.5); 

to_string is a new library function introduced in N1803 (r0), N1982 (r1) and N2408 (r2) \"Simple Numeric Access\". There are also the stod function to perform the reverse operation.

If you do want to have a different output format than \"%f\", use the snprintf or ostringstream methods as illustrated in other answers.

回答4:

If you use C++, avoid sprintf. It\'s un-C++y and has several problems. Stringstreams are the method of choice, preferably encapsulated as in Boost.LexicalCast which can be done quite easily:

template <typename T>
std::string to_string(T const& value) {
    stringstream sstr;
    sstr << value;
    return sstr.str();
}

Usage:

string s = to_string(42.5);

回答5:

You can use std::to_string in C++11

double d = 3.0;
std::string str = std::to_string(d);

回答6:

sprintf is okay, but in C++, the better, safer, and also slightly slower way of doing the conversion is with stringstream:

#include <sstream>
#include <string>

// In some function:
double d = 453.23;
std::ostringstream os;
os << d;
std::string str = os.str();

You can also use Boost.LexicalCast:

#include <boost/lexical_cast.hpp>
#include <string>

// In some function:
double d = 453.23;
std::string str = boost::lexical_cast<string>(d);

In both instances, str should be \"453.23\" afterward. LexicalCast has some advantages in that it ensures the transformation is complete. It uses stringstreams internally.

回答7:

I would look at the C++ String Toolkit Libary. Just posted a similar answer elsewhere. I have found it very fast and reliable.

#include <strtk.hpp>

double pi = M_PI;
std::string pi_as_string  = strtk::type_to_string<double>( pi );

回答8:

Herb Sutter has an excellent article on string formatting. I recommend reading it. I\'ve linked it before on SO.

回答9:

The problem with lexical_cast is the inability to define precision. Normally if you are converting a double to a string, it is because you want to print it out. If the precision is too much or too little, it would affect your output.

回答10:

You could also use stringstream.

回答11:

Heh, I just wrote this (unrelated to this question):

string temp = \"\";
stringstream outStream;
double ratio = (currentImage->width*1.0f)/currentImage->height;
outStream << \" R: \" << ratio;
temp = outStream.str();

/* rest of the code */

回答12:

You may want to read my prior posting on SO. (Macro\'ed version with a temporary ostringstream object.)

For the record: In my own code, I favor snprintf(). With a char array on the local stack, it\'s not that inefficient. (Well, maybe if you exceeded the array size and looped to do it twice...)

(I\'ve also wrapped it via vsnprintf(). But that costs me some type checking. Yelp if you want the code...)

回答13:

Take a look at sprintf() and family.

回答14:

Normaly for this operations you have to use the ecvt, fcvt or gcvt Functions:

/* gcvt example */
#include <stdio.h>
#include <stdlib.h>

main ()
{
  char buffer [20];
  gcvt (1365.249,6,buffer);
  puts (buffer);
  gcvt (1365.249,3,buffer);
  puts (buffer);
  return 0;
}

Output:
1365.25
1.37e+003   

As a Function:

void double_to_char(double f,char * buffer){
  gcvt(f,10,buffer);
}

回答15:

You could try a more compact style:

std::string number_in_string;

double number_in_double;

std::ostringstream output;

number_in_string = (dynamic_cast< std::ostringstream*>(&(output << number_in_double <<

std::endl)))->str(); 

回答16:

Use to_string().
example :

#include <iostream>   
#include <string>  

using namespace std;
int main ()
{
    string pi = \"pi is \" + to_string(3.1415926);
    cout<< \"pi = \"<< pi << endl;

  return 0;
}

run it yourself : http://ideone.com/7ejfaU
These are available as well :

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

回答17:

You can convert any thing to anything using this function:

template<class T = std::string, class U>
T to(U a) {
    std::stringstream ss;
    T ret;
    ss << a;
    ss >> ret;
    return ret;
};

usage :

std::string str = to(2.5);
double d = to<double>(\"2.5\");