Exception not caught in try catch block

Question

I do a simple throw \"TEST THROW\" and it isn\'t caught in my catch (std::exception& e). Is it because I\'m catching an std::exception& e?

Answer 1

This also may happen when you throw an exception of a an inherted type but the inhertence is private

Answer 2

You can add a catch (...) block to get it.

Answer 3

Since this is not a MCVE (what is Core?), I can't address explicitly the problem, but you are surely missing to

#include <exception>

Actually, GCC compiles even without the inclusion, but exception won't be caught and you will end up with

terminate called after throwing an instance of 'std::exception'

what(): std::exception

./{program}: {PID} Aborted (core dumped)

Answer 4

You're throwing a const char*. std::exception only catches std::exception and all derived classes of it. So in order to catch your throw, you should throw std::runtime_error("TEST THROW") instead. Or std::logic_error("TEST THROW"); whatever fits better. The derived classes of std::exception are listed here.

Answer 5

Another reason people may hit this issue, especially if they've been writing Java recently, is that they may be throwing a pointer to the exception.

/* WARNING WARNING THIS CODE IS WRONG DO NOT COPY */
try {
    throw new std::runtime_error("catch me");
} catch (std::runtime_error &err) {
    std::cerr << "exception caught and ignored: " << err.what() << std::end;
}
/* WARNING WARNING THIS CODE IS WRONG DO NOT COPY */

will not catch the std::runtime_error* you threw. It'll probably die with a call to std::terminate for the uncaught exception.

Don't allocate the exception with new, just throw the constructor by-value, e.g.

try {
    /* note: no 'new' here */
    throw std::runtime_error("catch me");
} catch (std::runtime_error &err) {
    std::cerr << "exception caught and ignored: " << err.what() << std::end;
}