Reputation: 64266
I'm trying to convert std::string
to float/double
.
I tried:
std::string num = "0.6";
double temp = (double)atof(num.c_str());
But it always returns zero. Any other ways?
Upvotes: 119
Views: 350631
Reputation: 6484
With C++17, you can use std::from_chars
, which is a lighter weight faster alternative to std::stof
and std::stod
. It doesn't involve any memory allocation or look at the locale, and it is non-throwing.
The std::from_chars
function returns a value of type from_chars_result
, which is basically a struct with two fields:
struct from_chars_result {
const char* ptr;
std::errc ec;
};
By inspecting ec
we can tell if the conversion was successful:
#include <iostream>
#include <charconv>
int main()
{
const std::string str { "12345678901234.123456" };
double value = 0.0;
auto [p, ec] = std::from_chars(str.data(), str.data() + str.size(), value);
if (ec != std::errc()) {
std::cout << "Couldn't convert value";
}
return 0;
}
NB: you need a fairly up-to-date compiler (e.g. gcc11) for std::from_chars
to work with floating point types.
Upvotes: 4
Reputation: 16091
The Standard Library (C++11) offers the desired functionality with std::stod
:
std::string s = "0.6"
std::wstring ws = "0.7"
double d = std::stod(s);
double dw = std::stod(ws);
Generally for most other basic types, see <string>
. There are some new features for C strings, too. See <stdlib.h>
Upvotes: 131
Reputation: 4497
My Problem:
My solution (uses the Windows function _wcstod_l):
// string to convert. Note: decimal seperator is ',' here
std::wstring str = L"1,101";
// Use this for error detection
wchar_t* stopString;
// Create a locale for "C". Thus a '.' is expected as decimal separator
double dbl = _wcstod_l(str.c_str(), &stopString, _create_locale(LC_ALL, "C"));
if (wcslen(stopString) != 0)
{
// ... error handling ... we'll run into this because of the separator
}
HTH ... took me pretty long to get to this solution. And I still have the feeling that I don't know enough about string localization and stuff...
Upvotes: 0
Reputation: 4449
If you don't want to drag in all of boost, go with strtod(3)
from <cstdlib>
- it already returns a double.
#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
using namespace std;
int main() {
std::string num = "0.6";
double temp = ::strtod(num.c_str(), 0);
cout << num << " " << temp << endl;
return 0;
}
Outputs:
$ g++ -o s s.cc
$ ./s
0.6 0.6
$
Why atof() doesn't work ... what platform/compiler are you on?
Upvotes: 8
Reputation: 1483
You don't want Boost lexical_cast for string <-> floating point anyway. That subset of use cases is the only set for which boost consistently is worse than the older functions- and they basically concentrated all their failure there, because their own performance results show a 20-25X SLOWER performance than using sscanf and printf for such conversions.
Google it yourself. boost::lexical_cast can handle something like 50 conversions and if you exclude the ones involving floating point #s its as good or better as the obvious alternatives (with the added advantage of being having a single API for all those operations). But bring in floats and its like the Titanic hitting an iceberg in terms of performance.
The old, dedicated str->double functions can all do 10000 parses in something like 30 ms (or better). lexical_cast takes something like 650 ms to do the same job.
Upvotes: 0
Reputation: 328
I had the same problem in Linux
double s2f(string str)
{
istringstream buffer(str);
double temp;
buffer >> temp;
return temp;
}
it works.
Upvotes: 3
Reputation: 11
As to why atof()
isn't working in the original question: the fact that it's cast to double makes me suspicious. The code shouldn't compile without #include <stdlib.h>
, but if the cast was added to solve a compile warning, then atof()
is not correctly declared. If the compiler assumes atof()
returns an int, casting it will solve the conversion warning, but it will not cause the return value to be recognized as a double.
#include <stdlib.h>
#include <string>
...
std::string num = "0.6";
double temp = atof(num.c_str());
should work without warnings.
Upvotes: 1
Reputation: 14459
Rather than dragging Boost into the equation, you could keep your string (temporarily) as a char[]
and use sprintf()
.
But of course if you're using Boost anyway, it's really not too much of an issue.
Upvotes: 0
Reputation: 8931
The C++ 11 way is to use std::stod and std::to_string. Both work in Visual Studio 11.
Upvotes: 2
Reputation: 9552
You can use std::stringstream:
#include <sstream>
#include <string>
template<typename T>
T StringToNumber(const std::string& numberAsString)
{
T valor;
std::stringstream stream(numberAsString);
stream >> valor;
if (stream.fail()) {
std::runtime_error e(numberAsString);
throw e;
}
return valor;
}
Usage:
double number= StringToNumber<double>("0.6");
Upvotes: 14
Reputation: 4569
Yes, with a lexical cast. Use a stringstream and the << operator, or use Boost, they've already implemented it.
Your own version could look like:
template<typename to, typename from>to lexical_cast(from const &x) {
std::stringstream os;
to ret;
os << x;
os >> ret;
return ret;
}
Upvotes: 11
Reputation: 635
double myAtof ( string &num){
double tmp;
sscanf ( num.c_str(), "%lf" , &tmp);
return tmp;
}
Upvotes: 2
Reputation: 8447
std::string num = "0.6";
double temp = ::atof(num.c_str());
Does it for me, it is a valid C++ syntax to convert a string to a double.
You can do it with the stringstream or boost::lexical_cast but those come with a performance penalty.
Ahaha you have a Qt project ...
QString winOpacity("0.6");
double temp = winOpacity.toDouble();
Extra note:
If the input data is a const char*
, QByteArray::toDouble
will be faster.
Upvotes: 142
Reputation: 44804
This answer is backing up litb in your comments. I have profound suspicions you are just not displaying the result properly.
I had the exact same thing happen to me once. I spent a whole day trying to figure out why I was getting a bad value into a 64-bit int, only to discover that printf was ignoring the second byte. You can't just pass a 64-bit value into printf like its an int.
Upvotes: 1
Reputation: 79790
You can use boost lexical cast:
#include <boost/lexical_cast.hpp>
string v("0.6");
double dd = boost::lexical_cast<double>(v);
cout << dd << endl;
Note: boost::lexical_cast throws exception so you should be prepared to deal with it when you pass invalid value, try passing string("xxx")
Upvotes: 7
Reputation: 81926
Lexical cast is very nice.
#include <boost/lexical_cast.hpp>
#include <iostream>
#include <string>
using std::endl;
using std::cout;
using std::string;
using boost::lexical_cast;
int main() {
string str = "0.6";
double dub = lexical_cast<double>(str);
cout << dub << endl;
}
Upvotes: 31