c++ - 日志记录，如何获得命令结束？

Question

所以我用这样的Log class :

#include <stdio.h>
#include <iostream>

class Log
{
public:
    int i;
    Log()
    {
        i = 0;
    }

    template <class T>
    Log &operator<<(const T &v)
    {
        i++;
        std::cout << i << ":"  << v << ";" <<std::endl;
        return *this;
    }
    Log &operator<<(std::ostream&(*f)(std::ostream&)) 
    {
        i++;
        std::cout << i << ":"  << *f << ";" <<std::endl;
        return *this;
    }

    ~Log()
    {
        std::cout << " [end of message]" << std::endl;
    }
};

我这样使用:

#include <log.h>

int main()
{
    Log a;
    a << "here's a message" << std::endl;
    a << "here's one with a number: " << 5;
    std::cin.get();
}

我希望我的日志类在我输入“;”时得到意思是如果我有 a << "here's a message" << std::endl;我希望它能够得到它是 oue 日志消息和 a << "here's one with a number: " << 5;是另一个。

最近它输出下一条消息:

1:here's a message;
2:
;
3:here's one with a number: ;
4:5;

我想保留它的语法(无限数量的 << ，大范围的值类型，在 api 中没有 ( 和 ) )但让它输出:

1:here's a message
;
2:here's one with a number: 5;

如何做这样的事情？

Answer 1

制作operator<<返回一个临时值，它将放置 endl销毁并转发所有operator<<调用主要对象。这样，endl保证只调用一次。

class Log
{
struct EndlOnDeath {
    Log* log;
    EndlOnDeath(Log* ptr)
        : log(ptr) {}
    template<typename T> EndlOnDeath& operator<<(const T& val) {
        (*log) << val;
    }
    ~EndlOnDeath() {
        (*log) << std::endl;
    }
};

public:
    int i;
    Log()
    {
        i = 0;
    }

    template <class T>
    EndlOnDeath operator<<(const T &v)
    {
        i++;
        std::cout << i << ":"  << v << ";";
        return this;
    }
    Log &operator<<(std::ostream&(*f)(std::ostream&)) 
    {
        i++;
        std::cout << i << ":"  << *f << ";" <<std::endl;
        return *this;
    }

    ~Log()
    {
        std::cout << " [end of message]" << std::endl;
    }
};