What is the scope of the exception object in C++? does it go out of scope as soon as catch handler is executed? Also, if I create an unnamed exception object and throw it, then while catching that exception does it matter if I catch it by const reference or a non-const reference?
When a throw
expression is evaluated, an exception object is initialized from the value of the expression. The exception object which is thrown gets its type from the static type of the throw expression ignoring any const
and volatile
qualifiers. For class types this means that copy-initialization is performed.
The exception object's scope is outside of the scope of the block where the throw occurs. Think of it as living in a special exception area off to one side of the normal call stack where local objects live.
Inside a catch
block, the name initialized with the caught exception object is initialized with this exception object and not the argument to throw
, even if this was an lvalue.
If you catch
via non-const reference, then you can mutate the exception object, but not what it was initialized from. You can alter the behaviour of the program if you re-throw the exception in ways that you couldn't if you caught by value or const reference (const_cast
s aside).
The exception object is destroyed when the last catch block that does not exit via a re-throw (i.e. a parameterless throw expression evaluation) completes.
catch(...) { std::exception *p = nullptr; try { throw; } catch (const std::exception &e) { p = &e; } catch(...) {} if (p) { std::cerr << p->what() << std::endl; } }
Is this valid code? The inner catch doesn't exits via a re-throw. Is the outer catch(...) the last block? –
Erda The exception object is available only in catch
block. You cannot use the exception object outside the catch
block. Following steps happen when you throw an exception and catch:
try
{
MyException anObject;
throw anObject; //1
}
catch(MyException exObject)
{
}
- The
throw
clause (//1) receives the local objectanObject
, and treats it as a value argument: it creates a copy of theanObject
. - the
catch
handler catches an MyException Object,which again is a value parameter. At this moment another copy is created. - If the
catch
handler would have implemented so as to receive a reference to an object(catch (MyException &o))
, the second copy is avoided. - if
catch
handler receives the exception object byconst&
then you can only callconst
methods.
throw
statement is a pretty bad idea, precisely because that could lead to a redundant copy. Do throw Myexception();
instead. Apart from that, if you declare an exception object with a name (as in this answer) then of course it can be accessed outside the catch
block in the scope it was declared in — in this case, inside the try
block. –
Meso First of all, the object you throw goes out of scope almost immediately. What's going to be caught by exception handlers is a copy of original object. That copy will be deleted after catch handler is executed unless you catch it by value (not by reference). In this case there will be another copy created. But you should catch it by reference (preferably const one) anyway.
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