如何轻松地使std :: cout线程安全？ [英] How to easily make std::cout thread-safe?

问题描述

我有一个多线程的应用程序，它大量使用 std :: cout 用于没有任何锁定的日志。在这种情况下，如何轻松添加锁机制，使std :: cout线程安全？

I have a multi-threaded application, which heavily uses std::cout for logging without any locking. In such a case, how to easily add lock mechanism to make std::cout thread-safe?

我不想搜索每次出现的 std :: cout 并添加一行锁定代码。太麻烦了。

I don't want to search for each occurrence of std::cout and add a line of locking code. That is too tedious.

任何更好的做法？

推荐答案

猜测你可以实现你自己的类，它包装 cout 并将互斥体与它关联。该类的运算符<< 将执行三项操作：

I guess you could implement your own class which wraps cout and associates a mutex with it. The operator << of that new class would do three things:

1. 为互斥创建一个锁，可能阻塞其他线程


2. 做输出，即对包装执行操作符<< 流和传递的参数


3. 构造一个不同类的实例，将锁传递给

1. create a lock for the mutex, possibly blocking other threads
2. do the output, i.e. do the operator << for the wrapped stream and the passed argument
3. construct an instance of a different class, passing the lock to that

这个不同的类将保持锁和委托运算符< 到包装的流。第二个类的析构函数最终会破坏锁并释放互斥体。

This different class would keep the lock and delegate operator << to the wrapped stream. The destructor of that second class would eventually destroy the lock and release the mutex.

所以任何输出你写为单个语句，即作为单个序列<< 调用，只要所有输出都通过具有相同互斥量的对象，就会以原子方式打印。

So any output you write as a single statement, i.e. as a single sequence of << invocations, will be printed atomically as long as all your output goes through that object with the same mutex.

让我们调用两个类 synchronized_ostream 和 locked_ostream 。如果 sync_cout 是 synchronized_ostream 的实例，其中包含 std :: cout ，然后序列

Let's call the two classes synchronized_ostream and locked_ostream. If sync_cout is an instance of synchronized_ostream which wraps std::cout, then the sequence

sync_cout << "Hello, " << name << "!" << std::endl;

会导致以下操作：

1. synchronized_ostream :: operator< 会获取锁


2. synchronized_ostream :: operator<< 将委托打印Hello到 cout


3. operator<<<<（std :: ostream& const char *）会输出Hello。


4. synchronized_ostream :: operator<< 会构造一个 locked_ostream ，并将锁传递给


5. locked_ostream :: operator<< 会将 name 的打印委托给 cout


6. 运算符<<（std :: ostream& std :: string）会输出名称


7. cout 的同一授权发生在感叹号和结尾操纵符


8. locked_ostream 临时获得破坏，锁被释放

1. synchronized_ostream::operator<< would aquire the lock
2. synchronized_ostream::operator<< would delegate the printing of "Hello, " to cout
3. operator<<(std::ostream&, const char*) would print "Hello, "
4. synchronized_ostream::operator<< would construct a locked_ostream and pass the lock to that
5. locked_ostream::operator<< would delegate the printing of name to cout
6. operator<<(std::ostream&, std::string) would print the name
7. The same delegation to cout happens for the exclamation point and the endline manipulator
8. The locked_ostream temporary gets destructed, the lock is released

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