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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?
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.
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>
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;
}
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");
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;
}
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")
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?
I had the same problem in Linux
double s2f(string str)
{
istringstream buffer(str);
double temp;
buffer >> temp;
return temp;
}
it works.
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.
double myAtof ( string &num){
double tmp;
sscanf ( num.c_str(), "%lf" , &tmp);
return tmp;
}
The C++ 11 way is to use std::stod and std::to_string. Both work in Visual Studio 11.
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.
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.
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.
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.
My Problem:
Locale independent string to double (decimal separator always '.')
Error detection if string conversion fails
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...
In C++, how do I get ALL of the text after a space. I am trying to make my own coding language, so I want the user to be able to enter (/print (text here)) and print the text the user has entered. I want this to be all in one line; without having the user to input the command, then input the thing they want to output. Thank you to anyone who replies in advance.
Try this way. It will give you all the characters after the first space in the string.
std::string x = "ABC CDEFG HIJKL";
x.substr(x.find(" ") + 1);
Leveraging <algorithm>
The following will work with C++11:
#include <string>
#include <algorithm>
#include <cctype>
#include <iostream>
#include <iterator>
bool is_blank(char ch)
{
return std::isblank(static_cast<unsigned char>(ch));
}
int main() {
std::string inp = "print foo";
auto it = std::find_if(inp.begin(), inp.end(), is_blank);
it = std::find_if_not(it, inp.end(), is_blank);
std::copy(it, inp.end(), std::ostream_iterator<char>(std::cout));
}
Run this code in Compiler Explorer.
Note that we're only iterating over the input string once. Also note that this solution leverages the algorithms which come with the C++ standard library - no raw loops required :-)
Using std::string's find functions
std::string has a ton of built-in functions. I'm pretty sure if C++ could be developed from scratch most of them wouldn't be there. But since we have them we put them to some use:
#include <string>
#include <algorithm>
#include <iostream>
#include <iterator>
int main() {
std::string inp = "print foo";
const std::string whitespace = " \t";
auto i = inp.find_first_of(whitespace);
i = inp.find_first_not_of(whitespace, i);
std::cout << inp.substr(i, inp.size() - i) << std::endl;
}
Run this code in Compiler Explorer.
I prefer the first solution since I find the last line a little more readable. std::copy might also be slightly more efficient. Here std::string::substr() returns a temporary string which gets destroyed once std::cout has printed it. Not ideal in terms of performance which might or might not matter here.
I have a vector of strings, like this:
{"abc"}{"def"}{"ghi"}
I want to concatenate them into a single string, with a separator like "-".
Is there a concise (pretty) way of doing this without using a typical for loop? I have c++03 and boost available to me.
Sure, boost provides a convenient algorithm for achieving what you are trying to do. In higher level languages you may have spotted a join function. Boost provides an equivalent algorithm in the join function.
#include <boost/algorithm/string/join.hpp>
using namespace std;
string data[] = {"abc","def","ghi"};
const size_t data_size = sizeof(data) / sizeof(data[0]);
vector<string> stringVector(data, data + data_size);
string joinedString = boost::algorithm::join(stringVector, "-");
Just for reference, there is currently a proposal for std::join, which you can check out here.
But since you have boost available, you can use boost::algorithm::join, which takes a sequence of strings and a separator, like so:
#include <iostream>
#include <string>
#include <vector>
#include <boost/algorithm/string/join.hpp>
int main() {
std::vector<std::string> words;
words.push_back("abc");
words.push_back("def");
words.push_back("ghi");
std::string result = boost::algorithm::join(words, "-");
std::cout << result << std::endl;
}
Prints:
abc-def-ghi
Another option using only the STL is:
std::ostringstream result;
if (my_vector.size()) {
std::copy(my_vector.begin(), my_vector.end()-1,
std::ostream_iterator<string>(result, "-"));
result << my_vector.back();
}
return result.str()
I'm trying to figure out how to print a string several times. I'm getting errors. I just tried the line:
cout<<"This is a string. "*2;
I expected the output: "This is a string. This is a string.", but I didn't get that. Is there anything wrong with this line? If not, here's the entire program:
#include <iostream>
using namespace std;
int main()
{
cout<<"This is a string. "*2;
cin.get();
return 0;
}
My compiler isn't open because I am doing virus scans, so I can't give the error message. But given the relative simplicity of this code for this website, I'm hoping someone will know if I am doing anything wrong by simply looking.
Thank you for your feedback.
If you switch to std::string, you can define this operation yourself:
std::string operator*(std::string const &s, size_t n)
{
std::string r; // empty string
r.reserve(n * s.size());
for (size_t i=0; i<n; i++)
r += s;
return r;
}
If you try
std::cout << (std::string("foo") * 3) << std::endl
you'll find it prints foofoofoo. (But "foo" * 3 is still not permitted.)
There is an operator+() defined for std::string, so that string + string gives stringstring, but there is no operator*().
You could do:
#include <iostream>
#include <string>
using namespace std;
int main()
{
std::string str = "This is a string. ";
cout << str+str;
cin.get();
return 0;
}
As the other answers pointed there's no multiplication operation defined for strings in C++ regardless of their 'flavor' (char arrays or std::string). So you're left with implementing it yourself.
One of the simplest solutions available is to use the std::fill_n algorithm:
#include <iostream> // for std::cout & std::endl
#include <sstream> // for std::stringstream
#include <algorithm> // for std::fill_n
#include <iterator> // for std::ostream_iterator
// if you just need to write it to std::cout
std::fill_n( std::ostream_iterator< const char* >( std::cout ), 2, "This is a string. " );
std::cout << std::endl;
// if you need the result as a std::string (directly)
// or a const char* (via std::string' c_str())
std::stringstream ss;
std::fill_n( std::ostream_iterator< const char* >( ss ), 2, "This is a string. " );
std::cout << ss.str();
std::cout << std::endl;
Indeed, your code is wrong.
C++ compilers treat a sequence of characters enclosed in " as a array of characters (which can be multibyte or singlebyte, depending on your compiler and configuration).
So, your code is the same as:
char str[19] = "This is a string. ";
cout<<str * 2;
Now, if you check the second line of the above snippet, you'll clearly spot something wrong. Is this code multiplying an array by two? should pop in your mind. What is the definition of multiplying an array by two? None good.
Furthermore, usually when dealing with arrays, C++ compilers treat the array variable as a pointer to the first address of the array. So:
char str[19] = "This is a string. ";
cout<<0xFF001234 * 2;
Which may or may not compile. If it does, you code will output a number which is the double of the address of your array in memory.
That's not to say you simply can't multiply a string. You can't multiply C++ strings, but you can, with OOP, create your own string that support multiplication. The reason you will need to do that yourself is that even std strings (std::string) doesn't have a definition for multiplication. After all, we could argue that a string multiplication could do different things than your expected behavior.
In fact, if need be, I'd write a member function that duplicated my string, which would have a more friendly name that would inform and reader of its purpose. Using non-standard ways to do a certain thing will ultimately lead to unreadable code.
Well, ideally, you would use a loop to do that in C++. Check for/while/do-while methods.
#include <iostream>
#include <string>
using namespace std;
int main()
{
int count;
for (count = 0; count < 5; count++)
{
//repeating this 5 times
cout << "This is a string. ";
}
return 0;
}
Outputs:
This is a string. This is a string. This is a string. This is a string. This is a string.
Hey there, I'm not sure that that would compile. (I know that would not be valid in c# without a cast, and even then you still would not receive the desired output.)
Here would be a good example of what you are trying to accomplish. The OP is using a char instead of a string, but will essentially function the same with a string.
Give this a whirl:
Multiply char by integer (c++)
cout<<"This is a string. "*2;
you want to twice your output. Compiler is machine not a human. It understanding like expression and your expression is wrong and generate an error .
error: invalid operands of types
'const char [20]' and 'int' to binary
'operator*'
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?
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.
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>
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;
}
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");
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;
}
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")
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?
I had the same problem in Linux
double s2f(string str)
{
istringstream buffer(str);
double temp;
buffer >> temp;
return temp;
}
it works.
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.
double myAtof ( string &num){
double tmp;
sscanf ( num.c_str(), "%lf" , &tmp);
return tmp;
}
The C++ 11 way is to use std::stod and std::to_string. Both work in Visual Studio 11.
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.
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.
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.
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.
My Problem:
Locale independent string to double (decimal separator always '.')
Error detection if string conversion fails
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...