Exception Handling - throw without type not allowed

[Alan B]

I've been caught out by this a couple of times now. I wanted to throw an exception if a function call failed but as I wasn't bothered about the reason for failure, more the robustness of the code, so I used the following :-

try
{
    if (!foo()) throw;
}
catch(...)
{

}

Problem is, throw without an object doesn't seem to get caught and causes the process to crash. My workaround was to use

#define THROW throw 1

and then call THROW whenever I needed to.

Can anyone explain why this fails as I thought this use of throw was legitimate ?

Alan

[Moved to C++ Gurus, added code tags ~kshegunov]

[mrjj]

Hi
Im not sure you are using it correctly.
The standard says:
"A throw-expression with no operand rethrows the exception being handled. The exception is reactivated with the existing temporary; no new temporary exception object is created. The exception is no longer considered to be caught; therefore, the value of uncaught_exception() will again be true."

So you are re-throwing a not existing exception and it seems not to be catchable due to missing operand.

[m.sue]

@Alan-B

I read: throw without argument is a re-throw of an exception you caught. So it is allowed only inside a catch clause.

-Michael.

[Alan B]

Many thanks for the replies. That clears it up.