与cout和printf的区别，而c ++中的多线程 [英] Difference with cout and printf while multithreading in c++

问题描述

一些背景：

我有一个使用pthreads的多线程的c ++程序。该程序是一个酒店预订系统，10名客人（每个自己的线程），一个值机桌（1个线程）和退房桌（1线程）。

I have a c++ program that is multithreaded using pthreads. The program is a hotel reservation system, with 10 guests (each their own thread), a check-in desk (1 thread) and a check-out desk (1 thread). There are only 5 rooms in the hotel that a guest can be in. I am using semaphores to enforce mutual exclusion and event ordering in this program.

问题：

这里是我的代码（只是需要的部分...）

here is my code (Just parts that are needed...)

sem_init(&openRooms, 0, 5);

sem_wait(&openRooms);   //waits for there to be an open room at the hotel

cout << "Guest " << guestNumber << " waits for check-in." << endl;

sem_wait(&checkInLine); //waits for an open spot at the check-in desk

有5位客人可以在一次的酒店，因为有5间房间。当我运行程序，我得到的输出（或类似...）

There are 5 guests that are able to be in the hotel at a time, because there are 5 rooms. When I run the program, I get the output (or something like...)

Guest Guest Guest Guest 24 waits for check-in. waits for check-in.1 waits for check-in. 
3 waits for check-in.

似乎cout是让多个打印同时运行，这就是为什么Guest在一行中多次打印。

It seems that cout is letting multiple prints be run at the same time, which is why "Guest" is printed multiple times in a row.

我试着用printf这样做，同样的问题不会发生。整个语句在另一个线程打印语句之前打印。

I have tried doing this with printf, and the same issue doesn't happen. The entire statement is printed before another thread can print the statement.

sem_wait(&checkInSt); //Only one person at check in receptionist at a time
printf("Guest %ld goes to the check-in receptionist.\n", my_rank);
currentThreadIn = my_rank; //Rank for receptionist to use in room assignment

输出：

Guest 1 waits for check in.
Guest 3 waits for check in.
Guest 2 waits for check in.
Guest 4 waits for check in.
Guest 0 waits for check in.

为什么会发生这种情况？是否有某种bug的cout使用？此外，有没有办法避免使用cout？我可以使用额外的信号量来确保cout语句在另一个可以打印之前打印，但我正在寻找一种不使用额外信号量的方法。

Why does this happen? Is there some sort of bugger that cout uses? Additionally, is there any way to avoid this using cout? I could use an additional semaphore to ensure that the cout statement prints before another can print, but I'm looking for a way without using that additional semaphore

推荐答案

printf("Guest %ld goes to the check-in receptionist.\n", my_rank);

当您使用 printf 格式化为内部缓冲区，然后通过单个操作（Guest 2等待签入）输出到控制台。

When you use printf, your string is formatted into an internal buffer and then output to the console in a single operation ("Guest 2 waits for check-in.").

cout << "Guest " << guestNumber << " waits for check-in." << endl;

当您使用 cout 输出到控制台多个部分 - Guest，其次是 guestNumber ，后跟 等待签入。，后跟 endl 。这是因为对<< 运算符的每次调用都像是一个单独的函数调用一样发生（它返回对同一 cout

When you use cout, your string is output to the console in multiple parts - "Guest", followed by guestNumber, followed by " waits for check-in.", followed by endl. This is because each call to the << operator takes place as if it were a separate function call (it returns a reference to the same cout object for the next call to use).

因此，虽然写入控制台本身是线程安全的和原子的，在 cout case它只对每个单独的子字符串使用原子。

So although writing to the console itself is thread-safe and atomic, in the cout case it's only atomic for each separate sub-string.

解决方案如果你不想使用 printf 将是a）使用信号量或其他锁定机制，或b）在使用 stringstream 打印之前格式化文本，然后将其作为单个字符串输出。

The solution if you don't want to use printf would be to a) use a semaphore or other locking mechanism, or b) format the text before printing it using stringstream, and then output it as a single string.

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