英文:
In C++, does a data race in one possible scenario render the program's behaviour undefined even if that scenario never gets executed?
问题
以下程序包含数据竞争：
#include <iostream>
#include <thread>
#include <chrono>
#include <string>
using namespace std::chrono_literals;
int sharedVar = 42;
void func(const std::string& threadName) {
int input;
while(true) {
std::cout << "[" << threadName << "]输入一个数字: " << std::endl;
std::cin >> input;
if(input == 42) {
sharedVar++;
}
else {
std::this_thread::sleep_for(1s);
}
}
}
int main() {
std::thread t(func, "线程");
t.detach();
func("主程序");
}
如果两个线程在没有同步的情况下访问sharedVar，并且用户没有提供input == 42，根据cppreference的说法，程序的行为是未定义的。
上面的程序是否保证可以正常工作（没有任何未定义的行为），如果用户从未提供input == 42？
还是说执行的情景（运行时是否提供42的输入）根本不重要，程序始终是未定义的。
我提出这个问题是因为有时在讨论数据竞争时，讨论会涉及执行的细节，例如在上面的情况下用户是否会在输入中提供42。
编辑：我想要添加我在网上找到的这篇文章，这篇文章似乎与此相关。根据我对这篇文章的理解，提供的输入并不重要。行为将始终是未定义的。
编辑2：基本上，我对此处提到的第2类函数很感兴趣。
英文:
The following program contains a data race:
#include <iostream>
#include <thread>
#include <chrono>
#include <string>
using namespace std::chrono_literals;
int sharedVar = 42;
void func(const std::string& threadName) {
int input;
while(true) {
std::cout<<"["<<threadName<<"]Enter a number: "<<std::endl;
std::cin>>input;
if(input == 42) {
sharedVar++;
}
else {
std::this_thread::sleep_for(1s);
}
}
}
int main() {
std::thread t(func, "thread");
t.detach();
func("main");
}
Both threads are accessing sharedVar without any synchronisation if they are given input == 42 from the user.
As per cppreference,
> If a data race occurs, the behavior of the program is undefined.
Is the above program guaranteed to work fine (without any undefined behaviour) if input == 42 is never provided?
or
Does the scenario of execution (whether 42 is provided in the input or not at runtime) not matter at all and the program is always undefined.
I ask this question because sometimes, during the discussion on data races, the discussion goes into the details of execution; e.g. in the above case whether the user will provide 42 or not in the input.
Edit
Just wanted to add this article I found online which seemed relevant. As per my understanding of this, the input which is given does not matter. The behaviour will remain undefined.
Edit 2
Basically the type 2 functions mentioned here is what I'm interested in.
答案1
得分: 1
以下是翻译好的部分：
"The following program contains a data race"（以下程序存在数据竞争）
"For some input values"（对于某些输入值）
"particularly both threads need to see input of 42, not just one thread"（特别是两个线程都需要看到输入为42，而不仅仅是一个线程）
"A data race is a property of an execution, not of the program in the abstract. From [intro.races#21]"（数据竞争是执行的属性，而不是抽象程序的属性。来自[intro.races#21]）
"If yes, then is the above program guaranteed to work fine (without any undefined behaviour) if input == 42 is never provided?"（如果是的话，那么上述程序是否保证在没有提供input == 42的情况下能够正常工作（没有任何未定义行为）？）
"Yes. It will continue to loop forever prompting for and accepting input."（是的。它将继续无限循环提示输入并接受输入。）
"Aside: I would be wary of supplying functions with reference parameters to std::thread::thread, although in this case it is fine, because you pass a const char[] not a rvalue std::string."（顺便说一下：我会谨慎提供引用参数给std::thread::thread函数，尽管在这种情况下是可以的，因为您传递的是const char[]而不是rvalue std::string。）
英文:
> The following program contains a data race
For some input values, particularly both threads need to see input of 42, not just one thread.
A data race is a property of an execution, not of the program in the abstract. From [intro.races#21]:
> The execution of a program contains a data race if
> If yes, then is the above program guaranteed to work fine (without any undefined behaviour) if input == 42 is never provided?
Yes. It will continue to loop forever prompting for and accepting input.
Aside: I would be wary of supplying functions with reference parameters to std::thread::thread, although in this case it is fine, because you pass a const char[] not a rvalue std::string.
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