Losing exception type when rethrowing an exception from a catch block

Question

Today I found a bug in a catch block:

catch (const exception& e){
    // do something
    // throw e; <-- bug!
    throw;    // <-- right thing to do
}

Basically if I rethrow the exception e explicitly, I get a new std::exception reconstructed, in fact the message from the what() method was the default std::string, instead of my custom built message.

What is the explanation? I thought that throw; is only a shorthand of throw ExceptionJustCaught;.

Answer 1

Exception objects are a bit special. They're constructed in a special place in memory, and their lifetime is determined by the catch block in which they're caught.

If you say throw e;, the lifetime of the original exception ends at the end of the catch block, and you are throwing a new exception by copying e, thus producing a classical slicing problem: Since e is a polymorphic reference to an object whose dynamical type is usually more-derived than std::exception, you end up slicing off the derived part of the object.

By contrast, throw; is a special statement that reactivates the original exception, so that it is no longer caught, and its lifetime does not end at the end of the block anymore. In fact, if you catch by non-constant reference, you can keep modifying the exception object and rethrow and thus communicate a state change up the lower catch blocks. But bear in mind that rethrowning is different from throwing a new exception!

Answer 2

Just a throw throws the current exception by reference. throw e copy constructs a new exception to throw. This is much the same way return works.