Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
As per this article:
If you try and lock a non-recursive mutex twice from the same thread without unlocking in between, you get undefined behavior.
My very naive mind tells me why don’t they just return an error? Is there a reason why this has to be UB?
Because it never happens in a correct program, and making a check for something that never happens is wasteful (and to make that check it needs to store the owning thread ID, which is also wasteful).
Note that it being undefined allows debug implementations to throw an exception, for example, while still allowing release implementations to be as efficient as possible.