StringStream

Ah, after I read carefully about stringstream. I have some code below:

Code:

---

int main(){
//string to Int
/* ******************************************** */
    stringstream ss;
    printf("FIRST TEST\n");
    string strNum;
    cin>>strNum;
    ss<<strNum;
    int a;
    ss>>a;
    a=a+1;
    cout<<a<<endl;  //NO PROBLEM HERE


    //int to String
/* ******************************************* */
    printf("SECOND TEST\n");
    int num;
    cin>>num;
    ss<<num;
    string str;
    ss>>str;
    str=str+" IS A NUMBER";
    cout<<str<<endl;  //PROBLEM: it just print "IS A NUMBER"


 /* ****************************************** */
    printf("THIRD TEST\n");
    stringstream ss_2;  string str_2;
    cin>>num;
    ss_2<<num;
    ss_2>>str_2;
    str_2=str_2+" IS A NUMBER";
    cout<<str_2<<endl;  //NORMAL AGAIN: it will print: "<num> IS A NUMBER"
    /* ****************** */
    return 0;
}

---

//Note: PROBLEM mean I don't want like that, not something's wrong :)

Uhm, in above code, at second test, if I use ss stringstream again, ss>>str will not put flow in ss into str. and the problem will be solved if I use another variable(example: ss_2). So, who can explain for me please :)

thanks :)

Try this:

Code:

---

#include <iostream>
#include <sstream>

int main(void){

        int num;
        std::stringstream ss;
        std::string output;

        std::cout << "Enter a number: "; //note NO PRINTF
        std::cin >> num;

        ss << num; //place num into stringstream
        ss >> output; //place contents of stringstream into string object

        output += " is a number"; //append string object
        std::cout << output << std::endl;

        return(0);
}

---

yes, I have tried it, But It meets same problem with me :(. this is my code, building from your code:

Code:

---

int main(){
int num;
    stringstream ss;
    string output;
    cout<<"Enter a number:";
    cin>>num;
    ss<<num;
    ss>>output;
    output+=" is a number";
    cout<<output<<endl;



    /* ********************** */
    cout<<"Enter a number again:";
    cin>>num;
    ss<<num;
    ss>>output;
    output+=" is a number";
    return(0);
}

---

If you want to reuse the stream then, before the line "ss << num;", you need a line "ss.clear();"

Quote:

---

So, who can explain for me please

---

I would but I cannot understand your post.

Just FYI, template functions can make arbitrary conversions fairly trivial:

Code:

---

#include <sstream>
#include <cctype>
#include <typeinfo>

template <typename Dst, typename Src>
bool iocpy(Dst& dst, const Src& src)
{
        bool success = true;
        std::stringstream io;
        if(!(io << src) || !(io >> dst))
                success = false;
        else
        {
                char ch;
                while(io.get(ch))
                        if(!std::isspace(ch))
                                success = false;
        }
        if(!success)
                dst = Dst();
        return success;
}

template <typename Dst, typename Src>
Dst iocpy(const Src& src)
{
        Dst dst;
        if(!iocpy(dst, src))
                throw std::bad_cast();
        return dst;
}

#include <iostream>

int main(int argc, char** argv)
{
        using namespace std;
        int sum = 0;
        for(int a = 1; a < argc; ++a)
        {
                try
                {
                        sum += iocpy<int>(argv[a]);
                }
                catch(bad_cast&)
                {       
                        cout << "Sorry, '" << argv[a] << "' is not an integer!" << endl;
                }                       
        }       
        cout << "Sum: " << sum << endl;
}

---

Quote:

---

Originally Posted by gardhr

Just FYI, template functions can make arbitrary conversions fairly trivial:

---

If you have Boost, I suggest using boost::lexical_cast, though it is less tolerant of whitespace. Incidentally, the first function template could also have been written as:

Code:

---

template <typename Dst, typename Src>
bool iocpy(Dst& dst, const Src& src)
{
    std::stringstream io;
    bool success = (io << src) && (io >> dst >> std::ws) && io.eof();
    if (!success)
    {
        dst = Dst();
    }
    return success;
}

---

I have use ss.clear() and nothing happened,yet. :(
@ laserlight and gardhg: I'm still reading your posts :D (because It's may strange for me, so I will read slowly), but can you give me STD library C++ solution, please. Because, I just can use it.(for contest purpose).
Of course, many solutions get, many ways to improve my knowledge :D

thanks :)

<sigh> It would also help if, in your last code fragment (post 3), you actually write output to std::cout so you can see what is happening. Writing to an object named "output" does not magically ensure anything is written to a file, cout, etc.

I am assuming that you have recompiled and linked your code, so you are actually executing it .....

@grumpy: your answer maybe useless to me, until I read my code again, and, :"> I haven't printed <string>output to monitor,yet.
And after that, I see that ss.clear() have solve my question completely.(at post 1 and post 3, too)

Again, I'm sorry sorry so much for my fault :(

By the way, one side effect of wrapping the conversion code in a function (template) is that the stringstream is used once for the conversion. This might be a little wasteful, but it means that you do not need to clear the state of the stream and reset it with an empty string in order to reuse the stringstream.