How to convert std::string to any of several types in constructor?

Question

I have the following class template which has a member variable whose type is determined by the template argument. I want to initialize the value of this member in the constructor, which only takes a std::string. Thus, my problem is that I need to convert std::string to any of several types (int, double, bool, string). I don't think I can specialize just the constructor, and I would rather not specialize the entire class for each type. The problem with my code below is that stringstream stops streaming out when it hits a space:

#include <iostream>
#include <string>
#include <sstream>

using namespace std;

template <typename Ty>
struct Test
{
    Ty value;

    Test(string str) {
        stringstream ss;
        ss.str(str);
        ss >> value;
    }
};


int main()
{
    Test<int> t1{"42"};
    Test<double> t2{"3.14159"};
    Test<string> t3{"Hello world"};

    cout << t1.value << endl << t2.value << endl << t3.value << endl;

    return 0;
}

The output of the above code is:

42
3.14159
Hello

instead of "Hello world". Is there some way to get stringstream to not stop at whitespace, or some other device that will do arbitrary conversions like I need?

I would recommend taking a look at Boost.Lexical_Cast

Benjamin Lindley
– Benjamin Lindley

2016-06-29 02:55:30 +00:00
Commented Jun 29, 2016 at 2:55 — Benjamin Lindley
– Benjamin Lindley, Commented Jun 29, 2016 at 2:55

JCall-Engineer · Accepted Answer · 2016-06-29 03:38:05Z

1

This works for me. Just declare a special implementation before the generalized implementation:

#include <iostream>
#include <string>
#include <sstream>

template<typename T>
struct Test {
    T value;
    Test(std::string);
};

template<>
inline Test<std::string>::Test(std::string str) {
    value = str;
}

template<typename T>
inline Test<T>::Test(std::string str) {
    std::stringstream ss;
    ss.str(str);
    ss >> value;
}

int main() {
    Test<int> t1{"42"};
    Test<double> t2{"3.14159"};
    Test<std::string> t3{"Hello world"};

    std::cout
        << t1.value << std::endl
        << t2.value << std::endl
        << t3.value << std::endl;

    return 0;
}

Here is an ideone working example.

edited Jun 29, 2016 at 3:38

answered Jun 29, 2016 at 3:09

JCall-Engineer

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3 Comments

Carlton Over a year ago

Hey, this works. I didn't know you could specialize the constructor without specializing the whole class. The key is not defining the constructor in the class at all; the general case and the specializations all go outside. On a side note, as a consequence of specializing outside of the class, I had to mark both versions of my constructor inline to avoid multiple definition errors.

JCall-Engineer Over a year ago

If it's defined in a header, yes you need to do that. I'll update my answer. Also realized that some compilers require template<> (perhaps with C++ 14) before the specialized definition, so updating that too

Carlton Over a year ago

Yes, GCC 4.9.3 required template<>. Forgot to mention that.

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