如何在 C++ 中将双精度数转换为字符串？
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How do I convert a double into a string in C++?
提问by Bill the Lizard
I need to store a double as a string. I know I can use printfif 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).
我需要将双精度存储为字符串。我知道printf如果我想显示它，我可以使用它，但我只想将它存储在一个字符串变量中，以便以后可以将它存储在地图中（作为value，而不是key）。
回答by Adam Rosenfield
// 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by Johannes Schaub - litb
The boost (tm)way:
该升压（TM）方式：
std::string str = boost::lexical_cast<std::string>(dbl);
The Standard C++way:
在标准C ++方式：
std::ostringstream strs;
strs << dbl;
std::string str = strs.str();
Note: Don't forget #include <sstream>
注意：不要忘记#include <sstream>
回答by kennytm
The Standard C++11way (if you don't care about the output format):
在标准C ++ 11的方式（如果你不关心输出格式）：
#include <string>
auto str = std::to_string(42.5);
to_stringis a new library function introduced in N1803(r0), N1982(r1) and N2408(r2) "Simple Numeric Access". There are also the stodfunction to perform the reverse operation.
to_string是N1803(r0)、N1982(r1) 和N2408(r2) “简单数字访问”中引入的新库函数。还有stod执行反向操作的功能。
If you do want to have a different output format than "%f", use the snprintfor ostringstreammethods as illustrated in other answers.
如果您确实希望"%f"使用与不同的输出格式，请使用其他答案中所示的snprintf或ostringstream方法。
回答by Yochai Timmer
You can use std::to_stringin C++11
您可以在 C++11 中使用std::to_string
double d = 3.0;
std::string str = std::to_string(d);
回答by Konrad Rudolph
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.LexicalCastwhich can be done quite easily:
如果您使用 C++，请避免sprintf. 它是非 C++y 并且有几个问题。Stringstreams 是首选的方法，最好像Boost.LexicalCast一样封装，这可以很容易地完成：
template <typename T>
std::string to_string(T const& value) {
stringstream sstr;
sstr << value;
return sstr.str();
}
Usage:
用法：
string s = to_string(42.5);
回答by coppro
sprintfis okay, but in C++, the better, safer, and also slightly slower way of doing the conversion is with stringstream:
sprintf没问题，但在 C++ 中，更好、更安全且速度稍慢的转换方式是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:
您还可以使用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, strshould be "453.23"afterward. LexicalCast has some advantages in that it ensures the transformation is complete. It uses stringstreams internally.
在这两种情况下，str都应该在"453.23"之后。LexicalCast 有一些优点，因为它确保转换是完整的。它stringstream在内部使用s。
回答by DannyK
I would look at the C++ String Toolkit Libary. Just posted a similar answer elsewhere. I have found it very fast and reliable.
我会看看C++ String Toolkit Libary。刚刚在其他地方发布了类似的答案。我发现它非常快速和可靠。
#include <strtk.hpp>
double pi = M_PI;
std::string pi_as_string = strtk::type_to_string<double>( pi );
回答by Fred Larson
Herb Sutter has an excellent article on string formatting. I recommend reading it. I've linked it beforeon SO.
回答by Fred Larson
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.
lexical_cast 的问题是无法定义精度。通常，如果您要将 double 转换为字符串，那是因为您想将其打印出来。如果精度太大或太小，都会影响您的输出。
回答by Firas Assaad
You could also use stringstream.
您也可以使用stringstream。
