Processing all passed overloads at once - c++

Question

I'm tired of making up on the spot debug codes and including <iostream> in every single file. So I wanted to make myself a universal, self-contained and lightweight debug class, that I would just include in the header, and forget.
I want to use something along the lines of
#include "debug.hpp"
debug DBG;
DBG << "foo and" << " bar";
//Or even better, just include it and do debug() << "foo and" << " bar";
So, I wrote this:
#include <iostream>
#include <string>
#include <chrono>
#include <ctime>
class Debug
{
public:
Debug &operator<<(std::string arg_0)
{
auto tempTime = std::chrono::system_clock::to_time_t(
std::chrono::system_clock::now() );
auto timeString(ctime(&tempTime));
timeString = timeString.substr(timeString.find(':') - 2, 8);
std::cout << timeString << " >> " << arg_0 << '\n';
return *this;
}
};
But of course, this doesn't work because, as I've learned, every overload operator causes this function (is it still called a function?) to trigger separately. Creating:
hour:minute:second >> foo and
hour:minute:second >> bar
Any way I could pass everything at once after the first overload operator appears? Maybe as a stringstream? Also, I won't be only passing strings, but anything that I need, will this require me to manually create a separate overload function for every signle type that I may pass?
P.S: Cross-plaform solution is optional, but welcome (Currently developing on Linux)

Answer 1

You may return an other class to do the job, something like:
class Helper
{
public:
~Helper() { std::cout << "\n"; }
template<typename T>
friend Helper&& operator << (Helper&&h, const T& t) {
std::cout << t;
return std::move(h);
}
};
class Debug
{
public:
template<typename T>
friend Helper operator<<(Debug&, const T& t)
{
auto tempTime = std::chrono::system_clock::to_time_t(
std::chrono::system_clock::now() );
auto timeString{ctime(&tempTime)};
timeString = timeString.substr(timeString.find(':') - 2, 8);
std::cout << timeString << " >> " << t;
return Helper{};
}
};

Answer 2

Each time you call operator<<, your code prints the time stamp and \n. And that's the problem. To avoid that, you can print the time stamp in the constructor of Debug, and print \n in the destructor.
class Debug {
public:
Debug() {
auto tempTime = std::chrono::system_clock::to_time_t(
std::chrono::system_clock::now() );
std::string timeString(ctime(&tempTime));
timeString = timeString.substr(timeString.find(':') - 2, 8);
std::cout << timeString;
}
~Debug() {
std::cout << "\n";
}
Debug &operator<<(std::string arg_0) {
std::cout << " >> " << arg_0;
return *this;
}
};
In order to debug types other than string, you make operator<< a template:
template <typename T>
Debug &operator<<(T &&arg_0) {
std::cout << " >> " << std::forward<T>(arg_0);
return *this;
}

Answer 3

I see 2 design problems here:
You try to create stream-like object. It means that it doesn't know, when the line ends, until you send EOL to it. Without this information, it doesn't know when to add prefix to "your" line and print it. Consider the two following situation:
DBG << "foo and" << " bar";
and
DBG << "foo and";
... (a lot of code) ...
DBG << " bar";
They look exactly the same inside your Debug class, because:
DBG << "foo and" << " bar"; == (DBG.operator<<("foo and")).operator<<(" bar");
And this is the same as:
DBG.operator<<("foo and");
DBG.operator<<("bar");
So you have to decide how to define the end of the message you want to print (and when do you want to measure the time: At the beginning or at the end of the message?).
When do you want to flush your stream? You have to send std::endl or std::flush to std::cout to flush it. Sending "\n" does not flush std::cout (this is important difference between std::endl and "\n"). If you do not flush it, it may be printed several minutes/hours later (it will wait in a buffer). On the other hand frequent buffer flushing may be a performance killer in application producing large amount of text.
Try to define how your stream should behave when you send to it "\n", std::endl and std::flush (std::endl should be converted to "\n"+std::flush).
About other questions:
I would use simple template to "transfer" parameter of operator<<() to std::cout. It would allow to use your class for any type that can be printed by std::cout. To make things simpler you can define the operator<<() outside your class, eg.:
template<typename tParam>
Debug &operator<<(Debug& stream, tParam const & myParam)
{
...
return stream;
}