I am not getting why if there is an active exception then if an exception is raised again, it leads to termination of program. Could someone explain?
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
What is it suppose to do? It can’t “double catch” or anything, nor does it make sense to simply ignore one. The standard specifies that if, during stack unwinding, another exception escapes, then
terminateshall be called.There is more discussion in the C++ FAQ. One “solution” is to wrap your destructor code in a try/catch block, and simply don’t let exceptions escape.
Another is to come up with some sort of custom exception chaining scheme. You’d do the above, but instead of ignoring an exception, you would append it to the currently thrown exception, and at the catch site handle both by hand.
The best solution, I think, it to try to remove the exceptional code from your destructor.