C++ -- Why 'what' print "Unknown exception" in catch scope?

Question

try
{
    range_error r("Hi I am hereeeee!");
    cout << r.what() << endl;   // print "Hi I am hereeeee!"  // case one
    exception *p2 = &r; 
    cout << p2->what() << endl; // print "Hi I am hereeeee!"  // case two
    throw p2;
}
catch (exception *e)
{
    cout << e->what() << endl;  // print "Unknown exception"  // case three
}

Question>

I don't know why case three prints "Unknown exception" instead of "Hi I am hereeeee!"? The printed result is copied from VS2010

score 5 · Accepted Answer · answered Aug 10 '11 at 21:30

5

This program results in undefined behavior. Because the variable r is declared inside the try block, it goes out of scope before the catch handler is invoked. At this point, e points to some area on the stack where an object of type range_error used to exist.

The following program should print the expected results:

range_error r("Hi I am hereeeee!");
try
{
    cout << r.what() << endl;   // print "Hi I am hereeeee!"  // case one
    exception *p2 = &r; 
    cout << p2->what() << endl; // print "Hi I am hereeeee!"  // case two
    throw p2;
}
catch (exception *e)
{
    cout << e->what() << endl;  // print "Hi I am hereeeee!"  // case three
}

However, you should not throw a pointer to an object, you should throw the object itself. The run-time library will store a copy of the range_error object and pass that copy to the exception handler.

Thus, you should use the following code instead:

try
{
    range_error r("Hi I am hereeeee!");
    cout << r.what() << endl;   // print "Hi I am hereeeee!"  // case one
    throw r;
}
catch (const exception& e)
{
    cout << e.what() << endl;  // print "Hi I am hereeeee!"  // case two
}

answered Aug 10 '11 at 21:30

André Caron

44,541
12
67
125

I should haven identified this problem myself. Thank you – q0987 Aug 10 '11 at 21:32
Jeez, you **must** have copied me, except your answer is better. – john Aug 10 '11 at 21:32
@John: Except, I also posted first, and fixed the comment in the `catch` block ;-) – André Caron Aug 10 '11 at 21:35
Apart from throwing by value, the generally used approach is to [catch by reference](http://stackoverflow.com/questions/2023032/catch-exception-by-pointer-in-c/2023045#2023045). The code in the answer is already like that, but I think it deserves an explicit remark. – noe Aug 10 '11 at 23:24
@ncases: indeed. However, the OP's been asking several questions about exception handling and seems well aware of [slicing](http://stackoverflow.com/questions/7014626/c-throw-dereferenced-pointer) objects in exception handlers. – André Caron Aug 10 '11 at 23:31

score 3 · Answer 2 · answered Aug 10 '11 at 21:29

3

Because by the time you get to the catch, your range_error has been destroyed and you're catching a dangling pointer. Either move the range_error declaration outside the try block or, better yet, throw an instance and catch by reference.

answered Aug 10 '11 at 21:29

Fred Larson

60,987
18
112
174

score 1 · Answer 3 · answered Aug 10 '11 at 21:31

Because the pointed to exception object has gone out of scope by the time you catch the exception. If you wrote

range_error r("Hi I am hereeeee!");
try
{
    cout << r.what() << endl;   // print "Hi I am hereeeee!"  // case one
    exception *p2 = &r; 
    cout << p2->what() << endl; // print "Hi I am hereeeee!"  // case two
    throw p2;
}
catch (exception *e)
{
    cout << e->what() << endl;  // print "Unknown exception"  // case three
}

it would print what you expected.

C++ -- Why 'what' print "Unknown exception" in catch scope?

3 Answers3