C ++包装2个串流(“对话”模仿)

时间:2013-04-27 00:44:47

标签: c++ stl stream operator-overloading wrapper

我正在为两个流写一个包装器。最后,我想模仿与Person的对话,以使我的代码清晰易用。

声明

class Person
{
public:
    void process();

private:
    std::stringstream _request;
    std::stringstream _response;
}

用法

Person daniel;
std::stringstream answer;
daniel << "Hello" << std::endl; // f.e std::endl means end of the phrase
daniel << "How" << "are you";
daniel << "doing?";
daniel >> answer;
daniel.process();
std::cout << "Daniel says: " << answer;
// or
std::cout << "Daniel says: " << daniel;

我觉得我可以通过重载<<>>运算符来实现它,但我被卡住了。我从来没有为这些目的使用过流。我google了很多,显然仍然无法理解基本的东西。

果壳

std::ostream& operator<<(std::ostream& o, const person& p)
{
    output << p._response.rdbuf();

    return (output);
}

std::istream& operator>>(std::istream& i, person& p)
{
    p._request << i.rdbuf();
    // internal process() call probably will be better
    return (p._request);
}

确实,有些例子会非常棒,但我会对那些让我摆脱僵局的粗略想法感到非常满意。

2 个答案:

答案 0 :(得分:2)

老实说,我不明白为什么你想要这个,但我想这是某种练习或家庭作业。无论如何,实现这个功能有一些细微之处,我怀疑你无法弄明白,所以这是工作解决方案。

#include <iostream>
#include <sstream>

class Person {
public:
    void process() {
        // Let's see what we've got in `_request`
        std::cout << _request.rdbuf() << std::endl;

        // Do some processing to produce correponding _response

        // In this example we hardcode the response
       _response << "I'm fine, thanks!";
    }

private:
    std::stringstream _request;
    std::stringstream _response;

    // Make them all friends so that they can access `_request` and `_response`
    friend Person& operator<<(Person& p, std::string const& s);
    friend Person& operator<<(Person& p, std::ostream& (*f)(std::ostream&));
    friend std::ostream& operator<<(std::ostream &o, Person& p);
    friend Person& operator>>(Person& p, std::string& s);
};

Person& operator<<(Person& p, const std::string& s) {
    p._request << s;

    return p;
}

// Notice this peculiar signature, it is required to support `std::endl`
Person& operator<<(Person& p, std::ostream& (*f)(std::ostream&)) {
    p._request << f;

    return p;                           
}

// Somewhat conventional stream operator overload (in terms of signature)
std::ostream& operator<<(std::ostream &o, Person& p) {
    o << p._response.rdbuf();

    return o;
}

Person& operator>>(Person& p, std::string& s) {
    // NOTE: This will read not the whole `_reponse` to `s`, but will stop on
    // the first whitespace. This is the default behavior of `>>` operator of
    // `std::stringstream`.
    p._response >> s;

    return p;
}

值得一提的是,在您尝试实现的功能方面,您的第一次尝试是完全错误的。这可以归结为这样一个事实:您似乎遵循了流操作符重载的传统教程,而这里的方法应该是不同的,以实现所需的功能。特别要注意建议的流操作符重载的签名。

此外,您必须添加更多重载以支持其他输入类型,例如int。从本质上讲,您必须添加更多类似于 std::basic_ostream 提供开箱即用的重载。特别要注意最后一个:

basic_ostream& operator<<(basic_ostream& st, 
                          std::basic_ostream& (*func)(std::basic_ostream&));

这家伙是支持std::endl的。请注意,我已经为您Person添加了类似的重载,以便std::endl正常工作(请参阅下文)。其他重载,比如原始类型,都可以作为练习。

阅读回应

你想要的那个。

int main() {
    Person daniel;

    daniel << "Hello" << std::endl;
    daniel << "How " << "are you";
    daniel << " doing?" << std::endl;

    // We are ready to process the request so do it
    daniel.process();

    // And Daniel's answer is...
    std::cout << "Daniel says: " << daniel << std::endl;

    return 0;
}

阅读回应:替代方式

这个有点笨拙和麻烦,基本上是Person& operator>>(Person& p, std::string& s)重载实施评论的结果(见上文)。

int main() {
    Person daniel;

    daniel << "Hello" << std::endl;
    daniel << "How " << "are you";
    daniel << " doing?" << std::endl;

    // We are ready to process the request so do it
    daniel.process();

    // And Daniel's answer is...
    std::string part1;
    std::string part2;
    std::string part3;

    // Will read "I'm"
    daniel >> part1;

    // Will read "fine,"
    daniel >> part2;

    // Will read "thanks!"
    daniel >> part3;

    std::cout << "Daniel says: " << part1 << " " << part2 << " " << part3 << std::endl;

    return 0;
}

答案 1 :(得分:1)

rdbuf()不返回字符串。你应该调用str()从stringstream获取一个std :: string。您需要将这两个函数声明为类的友元,以便访问其私有属性。

class Person
{
public:
    void process();

    friend std::ostream& operator<<(std::ostream& o, const person& p);
    friend std::istream& operator>>(std::istream& i, person& p);
private:
    std::stringstream _request;
    std::stringstream _response;
}

std::ostream& operator<<(std::ostream& o, const person& p)
{
    o << p._response.str();

    return (o);
}

std::istream& operator>>(std::istream& i, person& p)
{
    i >> p._request;
    // internal process() call probably will be better
    return i;
}