CODEWITHSUNDEEP
c++exception
user2436815
user2436815

Reputation: 3895

Meaning of 'throw()' in destructor signature

The following is a custom exception class from another person's c++ tutorial:

class MyException : public std::exception {
public:
    MyException(const std::string message) : m_message(message) { }
    ~MyException() throw() { }
    const std::string what() const throw() { return m_message.c_str(); }

private:
    std::string m_message;
}

I get most of what going on here,, except for the "throw()" right next to the destructor..

Does it mean throw() is called whenever the exception class is destructed..??

Upvotes: 0

Views: 475

Answers (2)

ComicSansMS
ComicSansMS

Reputation: 54589

This is an exception specification.

It means that the destructor is not supposed to throw any exception. If it attempts to throw one anyway, the program will invoke std::terminate (which will almost certainly crash the program). Note that not all compilers implement this behavior correctly (most notably, in VC++ throwing from a destructor declared throw() leads to unspecified behavior).

Note that exception specifications have been deprecated with C++11 in favor of noexcept and should no longer be used (for good reasons). Destructors are implicitly noexcept in C++11.

Upvotes: 5

Mark Ransom
Mark Ransom

Reputation: 308158

It's a declaration of what the function is allowed to throw. In this case, nothing.

See http://en.cppreference.com/w/cpp/language/except_spec

If you don't list an exception type inside the throw() and you later try to throw one from that function, you'll get an std::unexpected thrown instead.

If the throw() is left out as it usually is, then any exception may be thrown.

Upvotes: 2

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