C4297 warning in Visual Studio while using function-try-block (function assumed not to throw an exception but does)

问题

#include <exception>

struct FOO
{
  ~FOO() try
  {
    throw std::exception();
  }
  catch (...)
  {
    return; // Shall prevent the exception from being rethrown?
  }
};

Building this code in Visual Studio triggers C4297 warning (function assumed not to throw an exception but does).

Reaching the end of a catch clause for a function-try-block on a destructor also automatically rethrows the current exception as if by throw;, but a return statement is allowed. quoted from cppreference.com;

Do I interpret this sentence correctly? Does return from the catch statement shall prevent the exception from being rethrown?

I logged a bug but they closed it as duplicate. The other bug does not have a return statement but I think it makes all the difference.

Live example

回答1:

Do I interpret this sentence correctly? Does return from the catch statement shall prevent the exception from being rethrown?

I believe you are. For one, it is explicitly stated that in a constructor, the handler of a function-try-block may not include a return statement.

[except.handle]

13 If a return statement appears in a handler of the function-try-block of a constructor, the program is ill-formed.

The only way to explicitly leave such a handler is by throwing another exception. A return statement is disallowed precisely for the reason that it will swallow the exception. When we leave a handler implicitly, by flowing of the end

14 The currently handled exception is rethrown if control reaches the end of a handler of the function-try-block of a constructor or destructor. Otherwise, flowing off the end of the compound-statement of a handler of a function-try-block is equivalent to flowing off the end of the compound-statement of that function (see [stmt.return]).

The bit in [stmt.return] says that reaching the closing brace of a void returning function is equivalent to a return; at the end. So the first sentence tells us that in a handler of a destructor's function-try-block, flowing of the end is not a return;, it rethrows. There is no implicit return there.

This leaves only the conclusion that explicitly returning, by virtue of not being prohibited, must swallow the current exception.

来源：https://stackoverflow.com/questions/63176146/c4297-warning-in-visual-studio-while-using-function-try-block-function-assumed

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