What does mean this C++ code: "void Foo() throw;"?

Question

Question from the one interview.

Please explain what does this C++ code mean:

void Foo() throw;

Answer 1

void Foo() throw;

This is a syntax error. The grammar for exception specification (C++98 §15.4) is:

exception-specification:

throw ( type-id-list_opt )

Note that the parenthesis are required.

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On the other hand,

void Foo() throw();

means the function Foo() will not throw any exceptions.