What does a single "throw;" statement do? - c++

These days, I have been reading a lot the C++ F.A.Q and especially this page.
Reading through the section I discovered a "technique" that the author calls "exception dispatcher" that allows someone to group all his exception handling in one handy function:
void handleException()
{
try {
throw; // ?!
}
catch (MyException& e) {
//...code to handle MyException...
}
catch (YourException& e) {
//...code to handle YourException...
}
}
void f()
{
try {
//...something that might throw...
}
catch (...) {
handleException();
}
}
What bothers me is the single throw; statement: if you consider the given example then sure, it is obvious what it does: it rethrows the exception first caught in f() and deals with it again.
But what if I call handleException() on its own, directly, without doing it from a catch() clause ? Is there any specified behavior ?
Additionally for bonus points, is there any other "weird" (probably not the good word) use of throw that you know of ?
Thank you.

If you do a throw; on its own, and there isn't a current exception for it to rethrow, then the program ends abruptly. (More specifically, terminate() is called.)
Note that throw; is the only safe way to re-throw the current exception - it's not equivalent to
catch (exception const & e) { throw e; }

Yes, it specified behavior, it will call terminate;
15.1, para 8: If no exception is presently being handled, executing a
throw expression with no operand calls
terminate() (15.5.1).

That's so-called exception handler. It rethrows the "current exception" if any. If there's no exception currently being handled terminate() will be called.

Related

Rethrow from a function in the catch block

In this case, I have want to perform some actions based on the exception thrown and then re-throw the exception. Is this recommended - My aim is to do some work based on the exception thrown and rethrow it and have the application crash and generate dump that have the call stack in the exception.
class Foo
{
public:
void HandleException(const std::exception& ex)
{
// Log, report some metrics
throw;
}
void Work(//Some inputs)
{
try
{
// trying doing some work
}
catch (const std::exception& ex)
{
// This is really an exceptional situation, and the exception should be thrown which
// cause the exe to abort and create dump.
// Intention is to preserve call stack and have it in dump.
HandleException(ex);
}
}
}
Let me add another note to the question: When I have HandleException as a lambda function, throw in the lambda causes exception. Do I need to capture some state and how do I do that ?
When you catch an exception you have two options:
Achieve the original goal (contract) in some way, e.g. by retrying.
Report failure by throwing.
Rethrowing the original exception is one way to implement the second bullet point.
Yes that is valid. BUT it is also dangerous.
If you call throw; with no parameters and there is no exception currently in flight this will result in a call to std::terminate(). The problem with your current code is that anybody can call the function even when they are not in a catch block (which would result in termination).
So you may want to validate that an exception is propogating:
void HandleException(const std::exception& ex)
{
// Log, report some metrics
if (std::uncaught_exception()) {
throw;
}
// Maybe put an else here.
}
Also worth reading: GotW #47: Uncaught Exceptions

Is it safe to throw / catch on stack unwind?

Q: Is it safe to throw and catch an exception on stack unwind, or does the application call terminate on the second throw?
minimal example:
void some_function()
{
try
{
// do stuff here that can throw
throw std::runtime_error("blah");
} catch(const std::exception& re)
{
try // this code could be in some function called from here
{
// do something with re here that throws a logical_error
throw std::logical_error("blah blah"); // does this call terminate?
} catch(const std::logical_error& le)
{
}
}
}
I got curious after reading this question.
Note: I know you can/should catch(...) in a destructor, but does it make sense in general to have a try/catch in a catch block - maybe in some function called on the exception (re in my example)?
That's not really during stack unwinding. Once a catch block is entered, the stack has already been unwound.
And yes, that code is legal. See this question: Nested try...catch inside C++ exception handler?
Pubby's answer best answers the scenario you're describing.
As an addendum, while a stack is unwinding, the only user code that's executed is destructors (and the code those destructors call).
If you do throw in a destructor during this scenario, the standard specifies that std::terminate() will be called.

c++ exception handling

Learning "try & catch". What is wrong with the following code?
Thanks for the advice.
Error in execution:
terminate called without an active exception
Aborted
The code:
#include <stdio.h>
int main()
{
int a = 3;
try
{
if (a < 5)
throw;
}
catch (...)
{
printf ("captured\n");
}
return 0;
}
Your throw; statement tries to rethrow a current exception but there probably isn't one. You need something like
throw some_exception_object();
Inside of a try block, you have to specify what to throw. The only place you can use throw by itself is inside of a catch block to re-throw the current exception. If you call throw by itself without a current exception being active, you will kill your app, as you have already discovered.
Try this:
#include <stdio.h>
int main()
{
int a = 3;
try
{
if (a < 5)
throw 1; // throws an int
}
catch (...)
{
printf ("captured\n");
}
return 0;
}
You can throw anything you want, as long as you throw something.
There are four things, two major and two minor. One thing at a time...
1. Rethrow usage w/o active exception
A throw; statement is used to re-throw an exception that is currently caught. For example:
try {
do_something();
} catch (const std::exception &) {
throw; // This statement re-throws an exception that was caught in this "catch" block.
}
In your case, you are using throw; without catching any exceptions (in order words — it does not appear inside catch block directly or indirectly), thus your program is terminated. When there is a need to throw and not to re-throw an exception, like in your case, you must specify an exception object to be thrown. For example:
throw std::runtime_error("Something bad happened");
2. catch-all clause which does not re-throw a caught exception
Your catch-all clause (catch (...)) is perfectly legal C++. However, it does not re-throw caught exception. Even though it is a legal C++ code, such a usage is a taboo. C and C++ runtime is usually using special types of exceptions to implement certain functionality. For example, NPTL is using exceptions to implement a thread cancellation. If you catch that exception using catch (...), a thread won't be cancelled and you are going to have a bad time. Generally, you have to catch exceptions by their types. In almost all cases, exceptions are inherited from std::exception, and so you have to write catch (const std::exception &) or, if you expect to catch an exact type, - catch(const TypeYouExpect &). If you must, however, use catch-all, remember to re-throw. For example:
try {
do_something();
} catch (...) {
throw; // DO NOT FORGET TO RE-THROW.
}
3. Header naming...
You are including C header whereas C++ provides its own headers for standard C features. So, header:
#include <stdio.h>
.. should be:
#include <cstdio>
C++ specific C functions get special treatment. For example, they become available in std namespace. So that you can use std::open() instead of just open() or ::open(). No big deal, but is highly recommended way to go.
4. Return from main.
Unlike C, C++'s main() function is very special. It allows you not to have return 0;. This is a default behavior. So, unless you really need to return some value, you may save yourself some time by not typing return 0;. Remember, however, that main is the only function like that, and that everywhere else you must explicitly return something unless a function is marked void.
Hope it helps. Good Luck!
You need to actually throw some object. Even something as simple as
throw "error";
will catch the error like you want it to.
see it in action here
The statement to throw an exception is:
throw <expression>;
This statement:
throw;
is also called the re-throw statement and is use to re-throw an existing exception that has been caught. It is typically used in a catch block, for example, you look at the exception and decide if you can continue, retry or abort. In case you decide to abort, you re-throw the exception so that somebody else down the call stack will catch it and handle this error.
For example:
// getResult() calls can fail with a deadlock exception
// This method will retry up to 3 times before failing
Result getResultWithRetry()
{
int nbTry = 3;
for(;;) {
try {
return getResult();
} catch (DeadLockException& e) {
if (nbTry == 0) {
throw; // re-throw the deadlock exception
}
}
--nbTry;
}
}

Nested try-catch blocks?

int main ()
{
try
{
try
{
throw 5;
}
catch (int n)
{
throw;
}
}
catch (...)
{
cout << "Exception occurred";
}
}
This prints out "Exception occured" but
int main ()
{
try
{
try
{
throw;
}
catch (...)
{
throw;
}
}
catch (...)
{
cout << "Exception occurred";
}
}
This just errors. It seems like I'm doing the try-catch's exactly the same! The only difference is that in the first case I'm throwing an int, then a general exception, but in the second case, I'm throwing a general exception both times. Is the program confused as to which catch to go to?
There's no such thing as "general exception" and you throw no such thing.
In the first example, you throw an int, then you re-throw the exception that you are handling. That's the meaning of throw without an argument.
In the second example you start with an attempt to re-throw an exception that you are handling. As you are not handling an exception at that time, you get an error.
Your 2nd example terminates the program with a good error message (at least using g++ 4.6.1):
terminate called without an active exception
Aborted
This happens because you are trying to re-throw an exception, but since there are no active exception, the program terminates.
The current c++11 draft, in chapter 15.5.1 says this:
In some situations exception handling must be abandoned for less
subtle error handling techniques.
Then in the list of cases when std::terminate is called is this:
when a throw-expression with no operand attempts to rethrow an
exception and no exception is being handled (15.1),
So, the behavior from your 2nd example is well defined in the standard.

C++ catch constructor exception

I do not seem to understand how to catch constructor exception.
Here is relevant code:
struct Thread {
rysq::cuda::Fock fock_;
template<class iterator>
Thread(const rysq::cuda::Centers &centers,
const iterator (&blocks)[4])
: fock_()
{
if (!fock_) throw;
}
};
Thread *ct;
try { ct = new Thread(centers_, blocks); }
catch(...) { return false; } // catch never happens,
So catch statement do not execute and I get unhandled exception.
What did I do wrong? this is straight C++ using g++.
You have to throw an object, e.g.,
throw std::exception();
throw with no operand is only used inside of a catch block to rethrow the exception being handled by the catch block.
You have to throw something in order to catch anything.
Try changing the line
if (!fock_) throw;
to
if (!fock_) throw "";
and observe the difference.
You need to throw something. throw alone means to "re-throw" the current exception. If there is no current exception, unexpected gets called, which will probably abort your program.
It's best to pick a class from <stdexcept> that describes the problem. logic_error or a derivative to indicate programming mistakes, or runtime_error to denote exceptional conditions.