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Home/ Questions/Q 551785
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Editorial Team
Asked: May 13, 20262026-05-13T11:25:51+00:00 2026-05-13T11:25:51+00:00

std::exception class is defined as follows exception() throw() { } virtual ~exception() throw(); virtual

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std::exception class is defined as follows

exception() throw() { }
virtual ~exception() throw();
virtual const char* what() const throw();

what does the throw() syntax mean in a declaration?
Can throw() take parameters? What does no parameters mean?

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  1. Editorial Team
    Editorial Team
    2026-05-13T11:25:52+00:00Added an answer on May 13, 2026 at 11:25 am

    Without any parameter, it means that the mentioned functions does not throw any exceptions.

    If you specify anything as a parameter, you’re saying that the function will throw only exceptions of that type. Notice, however, this is not an enforcement to the compiler. If an exception of some other type happens to be thrown the program will call std::terminate().

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