CODEWITHSUNDEEP
c++multithreadingc++11mutex
Dakorn
Dakorn

Reputation: 903

How lock_guard<mutex> constructor can be compiled fine without mutex instance?

I am studying about threads in C++11 now, and I met the following line of code:

lock_guard<mutex> lg(mutex);

There is no variable mutex. mutex is only name of type.

Can anyone explain me how above line of code works?

Why compiler(GCC) doesn't print any error?

Complete code:

#include <iostream>
#include <thread>
#include <mutex>
#include <chrono>

using namespace std;

void do_something()
{
    lock_guard<mutex> lg(mutex);

    cout << "Working..." << endl;

    this_thread::sleep_for(chrono::milliseconds(3000));
}

int main()
{
    thread thd(do_something);
    thd.join();
}

Upvotes: 2

Views: 822

Answers (2)

Brandon Kohn
Brandon Kohn

Reputation: 1621

The compiler thinks this is a prototype function declaration:

lock_guard<mutex> lg(mutex);

To be clear, the compiler parses this as the declaration of a function named 'lg' which takes a mutex as a parameter and returns a lock_guard instance.

#include <mutex>

int main()
{
    using namespace std;
    lock_guard<mutex> lg(mutex);
    return 0;
}

vc12 output : warning C4930 : 'std::lock_guard<std::mutex> lg(std::mutex)' : prototyped function not called(was a variable definition intended ? )

Upvotes: 8

Some programmer dude
Some programmer dude

Reputation: 409196

In C++ structure, class, enumeration and union names are in their own namespace (not a C++ namespace), which allows you to have variables with the same name as a structure.

For example:

struct SomeStruct
{
    // Member...
};

SomeStruct SomeStruct;  // Valid declaration

As for you not getting an error, if the function you use the shown it in a member function, then it could be that the class has a member variable with the name mutex.

Upvotes: 4

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