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Comment by thestoicattack | Hacker News Reader

To avoid this insidious bug:

    std::lock_guard<std::mutex>(m);  // guard is immediately thrown away

versus the correct

    std::lock_guard<std::mutex> g(m);  // hold mutex til end of scope

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Note for people who don’t see the difference: if a guard doesn’t have a name, it won’t survive until the end of the scope, thus doesn’t guard anything.

I might be wrong here, but I think that the first one is a declaration of a guard with the name m using the default constructor.

The same way that int (a); and int a; are equivalent.

So the guard should remain until end of scope, but won't guard anything.

lock_guard doesn’t have a default ctor, you have to pass a mutex (or presumably invoke a move ctor, but I can’t see one on cppref).

Lock guard has no default constructor, so the line will in fact materialize a temporary lock guard around the mutex m, then throw it away at the semi-colon, thus locking nothing.

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