I've never tried using C++ exceptions until a few days ago, and I'm not sure if I'm doing this right. I'm trying to throw an exception on a constructor that fails, like
X::X() {
/*...*/
if(error)
throw;
/*...*/
}
And using it like:
try {
X a;
X b;
X c;
}
catch (...) {
// error handling
}
The debugger (VS) says there's an unhanded exception on the throw. Code after throw gets executed (I though throw worked like return), and the catch block isn't executed. What am I missing here?
I may have oversimplified this post a little bit, but the original code is a bit complex to post here.
It's because you do not throw an exception object. You should use throw like this:
throw std::runtime_error("Error message");
then you'll be able to catch it via
try {
...
} catch (const std::runtime_error& e) {
/* Handling */
}
Have a look at the stdexcept header.
You should throw some exception (generally an instance of some subclass of std::exception), e.g.
X::X() {
/*...*/
if(error)
throw std::runtime_error("my bad");
/*...*/
}
See std::runtime_error for more.
throw; without any exception makes only sense inside a catch block.
I agree with Danvil's answer: throwing an exception inside a constructor is poor taste.
Related
An exercise from C++ Primer asks
Why is it important that the what function [of exception classes] doesn’t throw?
Since there is no way to check my answer I was hoping to get an opinion. I thought possibly that it is an error (maybe terminate would've been called) to throw another exception during a catch clause (other than a rethrow throw;) while the current exception object is still being handled. It seems that is not the case though and it is completely okay to throw out of catch clauses:
#include <iostream>
using namespace std;
int main(){
try{
try{
throw exception();
} catch(exception err){ throw exception();}
} catch(exception err){ cout << "caught"} //compiles and runs fine, outputs "caught"
}
So program terminations are not a worry. It seems then, any problem that arises from what() throwing should, at the very least, be rectifiable by the user if they were so inclined.
Maybe then, the importance might be that while handling an error we do not want further unexpected errors to occur? Throws inside catch clauses are mainly intended for sending the exception object further up the call chain. A user may receive an error from deep in his program and does not want to worry that the error caught has to be associated with its own try block. Or maybe what() having its own throw may also lead to recursive effects (e.g. what() throws an exception, then we catch this exception and call what() but this then throws, and so on) meaning it might become impossible to handle any errors? How drastic can it be for what() to potentially throw?
I think there's nothing unclear - it's just as you described. If .what() method of an exception class throws an error, the whole catch effort was wasted:
try {
someDangerousOperation();
}
catch(std::exception e) {
// Ooops, instead of false,
//we get another exception totally unrelated to original error
someLogClassOrWhatever.save(e.what());
return false;
}
return true;
And Imagine the crazy code if you were expected to deal with what()'s exceptions:
try {
someDangerousOperation();
}
catch(std::exception e) {
// Not very fun
try {
someLogClassOrWhatever.save(e.what());
}
catch(...) {
alsoWhatHasFailedThatIsReallyGreat();
}
return false;
}
I think there's nothing more in that, probably the question is so simple it seems there must be some catch hiding in it. I think it's not the case.
std::exception::what() is noexcept. Consequently, if it throws, std::terminate is called. Yes, this is important.
Image a very curious coder with a slight tendency towards being a control freak (I know a couple of them myself), he really wants to know what is going wrong in his program and logs all errors with ex.what(). So he codes
try {
code();
}
catch(std::exception &e) {
std::cout<<e.what()
}
He is pretty pleased with the world in general and with himself in particular. But now it crosses his mind, that e.what() could throw an exception as well. So he is codes:
try{
try {
code();
}
catch(std::exception &e) {
std::cout<<e.what()
}
}
catch(std::exception &e) {
std::cout<<e.what()
}
A minute later he notices, that there is again an uncaught exception possible! Remember, he is a control freak, so he is going to write another try-catch block and than another and another
So you can bet any money, his project will be late - how could you do something like this to my friend? So please make sure e.what() doesn't throw:)
I guess it is the reason behind what being noexcept.
I need catch exceptions that are maybe thrown by code in try...catch construction, and do something depending on exception type. But also I want do some code if any exception are thrown.
I did so:
try
{
// code
}
catch (...)
{
try
{
throw;
}
catch (exc1)
{
// handling
}
catch (exc2)
{
// handling
}
// here is code that are executed if any exception are thrown
}
My code works, but I want to know whether the Standard says anything about it.
Maybe is there any better method?
UPD: Sorry for the slow answers to comments. Simple, English is a second language for me.
UPD: I think that code from AdelNick or second code from Steve Jessop or second code from David Rodríguez - dribeas are right for me.
Your approach is bad, because the // here is code that are executed if any exception are thrown part won't be executed if neither of exc1 or exc2 branch catch the exception. Your code is an equivalent version of the following:
try
{
// code
}
catch (exc1)
{
// handling
}
catch (exc2)
{
// handling
}
// here is code that are executed if any exception are thrown
The C++ language does not support a finally block, if that's what you're after. That is because an object's destructor is responsible for freeing resources, not a finally block. Here is Stroustrup's explanation of the topic.
You should write:
try
{
// code
}
catch (exc1)
{
// handling
}
catch (exc2)
{
// handling
}
catch (...)
{
// here is code that are executed if any *other* exception are thrown
}
If you want particular code to be executed as part of all three cases then you have two options: call it in all three places, or do something like this:
bool threw = true;
try
{
// code
threw = false;
}
catch (exc1)
{
// handling
}
catch (exc2)
{
// handling
}
catch (...) {
}
if (threw) {
// here is code that are executed if any exception are thrown
}
I would change the code order as following:
try
{
try
{
// code
}
catch (exc1)
{
// handling
throw; // edited. was: throw exc1;
}
catch (exc2)
{
// handling
throw; // edited. was: throw exc2;
}
}
catch(...)
{
// here is code that are executed if any exception are thrown
}
as it works if any other type of exception is thrown (not only exc1 or exc2). Also if the code common for all exceptions is the resource releasing, consider using RAII principle instead).
Looking at the comments it seems that you may be expecting something that won't happen, and that has already been discussed in other answers. But it could be a misunderstanding on our side. If the code inside the try can only throw one of exc1 or exc2 then yes, it will do what you expect.
From the comment update, it seems that the intention was to handle any exception, including those that are neither exc1 nor exc2. For that purpose, the code won't do what is expected and exc3 will propagate outside of the nested trys. Just flatten the two levels and add a catch (...) at the end.
From here on the answer was based on a false assumption about the question.
The pattern of rethrowing is supported by the standard and used commonly in situations where the handling of the exceptions in different contexts is the same to avoid code duplication:
void processException() {
// implement exception handling in all contexts
try {
throw;
} catch (exceptionType1 const & ex1) {
} catch (exceptionType2 const & ex2) {
// } catch (...) {
}
}
void f() {
try {
something();
} catch (...) {
processException();
}
}
void g() {
try {
anotherThing();
} catch (...) {
processException();
}
}
If the code to handle the exceptions is not going to be centralized for different functions, that pattern may complicate the code more than you need. You could consider a single try/catch with an extra control variable:
bool success = false;
try {
doSomething();
success = true;
} catch (exception1 const & ex1) {
...
} catch (exception2 const & ex2) {
...
// } catch (...) {
}
if (!success) {
// common code to all exceptions
}
I am using QT 4.8 (C++) for desktop application project, and writing exception handling which is as follows :
void callerMethod()
{
try
{
method1();
}
catch(Exception1& e)
{
// display critcal error message
// abort application
}
catch(std::Exception& e)
{
// print exception error message
}
catch(...)
{
// print unknown exception message
}
}
void method1()
{
try
{
// some initializations
// some operations (here exceptions can occur)
// clean-up code (for successful operation i.e no exception occurred)
}
catch(Exception1& e)
{
// clean-up code
throw e;
}
catch(Exception2& e)
{
// clean-up code
throw e;
}
catch(Exception3& e)
{
// clean-up code
throw e;
}
catch(...)
{
// clean-up code
throw;
}
}
So my question do I need to write the clean-up code in every catch block?
Is there any way I can avoid writing repeated code?
NOTE:: [ In method1() ] I want to re-throw exceptions which occurred
to my caller.So I can not catch them in single catch block,
because then type information will be lost.
Method1 can be much simplified by two concepts:
RAII. Put any clean-up code into destructors, and the clean-up code will be centralized.
Use the unqualified throw, and you won't need to know about the type of exception thrown.
So, method1() should look like:
void method1()
{
// some initializations of RAII objects
// some operations (here exceptions can occur)
}
The first catch clause in callerMethod can be removed if you derive Exception1 from std::exception, since the what() method is virtual.
You should throw exceptions as low as possible and catch them as high as possible in the call chain. This automatically leads to less code duplication, and centralizes error handling. You are throwing/catching all in one place, which seems a bit ... forced.
I often do this kind of thing (especially for program-ending exceptions:
int main()
try
{
function_calls_that_may_throw();
// ...
}
catch(my_exception& e)
{
e.do_exception_stuff();
}
catch(std::exception& e)
{
std::cout << e.what();
}
catch(...)
{
std::cout << "Something bad happened.\n";
}
This is only possible for throwing exceptions you don't plan on handling better or retrying the failed operation or something.
The pro of this approach is that all/most error handling code is at the top-level of your program, and all the functions in the call chain don't have to worry one bit about this stuff, all they do is throw an exception when they feel like it.
If all your clean up code is totally identical, you can do everything in your catch (...) block:
try {
// code
} catch (...) {
// cleanup
throw;
}
If your code varies slightly, you can always call a cleanup function:
try {
// code
} catch (exc1 ex) {
cleanup(args);
// exc1 specific
throw;
} catch (exc2 ex) {
cleanup(args);
// exc2 specific
throw;
} catch (...) {
cleanup(args);
throw;
}
Consider the following C++ code:
try {
throw foo(1);
} catch (foo &err) {
throw bar(2);
} catch (bar &err) {
// Will throw of bar(2) be caught here?
}
I would expect the answer is no since it is not inside the try block and I see in another question the answer is no for Java, but want to confirm C++ is also no. Yes, I can run a test program, but I'd like to know the language definition of the behavior in the remote case that my compiler has a bug.
No. Only exceptions thrown in the associated try block may be caught by a catch block.
No, It won't, An enclosing catch block upwards the hierarchy will be able to catch it.
Sample Example:
void doSomething()
{
try
{
throw foo(1);
}
catch (foo &err)
{
throw bar(2);
}
catch (bar &err)
{
// Will throw of bar(2) be caught here?
// NO It cannot & wont
}
}
int main()
{
try
{
doSomething();
}
catch(...)
{
//Catches the throw from catch handler in doSomething()
}
return 0;
}
No, a catch block handles the nearest exception, so if you try ... catch ( Exception &exc ) ... catch ( SomethingDerived &derivedExc ) the exception will be handled in the &exc block
You might achieve the desired behaviour by exception delegation to the calling method
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 ¢ers,
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.