Terminate called after throwing an instance of an exception, core dumped












7














I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:



 #include <iostream>

 #include <exception>



 class err : public std::exception

 {

 public:

      const char* what() const noexcept { return "error"; }

 };



 void f() throw()

 {

      throw err();

 }



 int main()

 {

      try

      {

           f();

      }

      catch (const err& e)

      {

           std::cout << e.what() << std::endl;

      }

 }


When I run it, I get the following runtime error:



 terminate called after throwing an instance of 'err'

   what():  error

 Aborted (core dumped)


If I move the try/catch logic completely to f(), i.e.



 void f() 

 {

      try

      {

           throw err();

      }

      catch (const err& e)

      {

            std::cout << e.what() << std::endl;

      }

 }


And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?






c++ c++11 exception







share|improve this question




















  • 5




    void f() throw() says that the function doesn't throw exceptions. And then you do.
    – Neil Butterworth
    1 hour ago










  • Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
    – Bathsheba
    1 hour ago






  • 1




    Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
    – Jesper Juhl
    1 hour ago


















7














I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:



 #include <iostream>

 #include <exception>



 class err : public std::exception

 {

 public:

      const char* what() const noexcept { return "error"; }

 };



 void f() throw()

 {

      throw err();

 }



 int main()

 {

      try

      {

           f();

      }

      catch (const err& e)

      {

           std::cout << e.what() << std::endl;

      }

 }


When I run it, I get the following runtime error:



 terminate called after throwing an instance of 'err'

   what():  error

 Aborted (core dumped)


If I move the try/catch logic completely to f(), i.e.



 void f() 

 {

      try

      {

           throw err();

      }

      catch (const err& e)

      {

            std::cout << e.what() << std::endl;

      }

 }


And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?






c++ c++11 exception







share|improve this question




















  • 5




    void f() throw() says that the function doesn't throw exceptions. And then you do.
    – Neil Butterworth
    1 hour ago










  • Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
    – Bathsheba
    1 hour ago






  • 1




    Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
    – Jesper Juhl
    1 hour ago
















7












7








7







I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:



 #include <iostream>

 #include <exception>



 class err : public std::exception

 {

 public:

      const char* what() const noexcept { return "error"; }

 };



 void f() throw()

 {

      throw err();

 }



 int main()

 {

      try

      {

           f();

      }

      catch (const err& e)

      {

           std::cout << e.what() << std::endl;

      }

 }


When I run it, I get the following runtime error:



 terminate called after throwing an instance of 'err'

   what():  error

 Aborted (core dumped)


If I move the try/catch logic completely to f(), i.e.



 void f() 

 {

      try

      {

           throw err();

      }

      catch (const err& e)

      {

            std::cout << e.what() << std::endl;

      }

 }


And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?






c++ c++11 exception







share|improve this question















I am going over C++ exceptions and am running into an error that I am unsure of why it is giving me issues:



 #include <iostream>

 #include <exception>



 class err : public std::exception

 {

 public:

      const char* what() const noexcept { return "error"; }

 };



 void f() throw()

 {

      throw err();

 }



 int main()

 {

      try

      {

           f();

      }

      catch (const err& e)

      {

           std::cout << e.what() << std::endl;

      }

 }


When I run it, I get the following runtime error:



 terminate called after throwing an instance of 'err'

   what():  error

 Aborted (core dumped)


If I move the try/catch logic completely to f(), i.e.



 void f() 

 {

      try

      {

           throw err();

      }

      catch (const err& e)

      {

            std::cout << e.what() << std::endl;

      }

 }


And just call it from main (without the try/catch block in main), then there is no error. Am I not understanding something, as it relates to throwing exceptions from functions?






c++ c++11 exception




c++ c++11 exception






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









Ron

10.5k21834




10.5k21834










asked 1 hour ago









HectorJ

474




474








  • 5




    void f() throw() says that the function doesn't throw exceptions. And then you do.
    – Neil Butterworth
    1 hour ago










  • Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
    – Bathsheba
    1 hour ago






  • 1




    Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
    – Jesper Juhl
    1 hour ago
















  • 5




    void f() throw() says that the function doesn't throw exceptions. And then you do.
    – Neil Butterworth
    1 hour ago










  • Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
    – Bathsheba
    1 hour ago






  • 1




    Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
    – Jesper Juhl
    1 hour ago










5




5




void f() throw() says that the function doesn't throw exceptions. And then you do.
– Neil Butterworth
1 hour ago




void f() throw() says that the function doesn't throw exceptions. And then you do.
– Neil Butterworth
1 hour ago












Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
– Bathsheba
1 hour ago




Absolutely! Are we only allowed to ask regex and database questions in this site? Have an upvote.
– Bathsheba
1 hour ago




1




1




Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
– Jesper Juhl
1 hour ago






Remove the throw(). You are saying "this function never throws" but then you violate that promise and throw anyway. That leads to std::terminate since you broke the rules of the language - at that point you cannot expect anything.
– Jesper Juhl
1 hour ago














1 Answer
1






active

oldest

votes


















10














The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.



Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.






share|improve this answer























  • Since c++11 no? unless I’ve lost my marbles.
    – Bathsheba
    1 hour ago










  • Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
    – HectorJ
    1 hour ago






  • 1




    @HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
    – François Andrieux
    1 hour ago













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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









10














The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.



Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.






share|improve this answer























  • Since c++11 no? unless I’ve lost my marbles.
    – Bathsheba
    1 hour ago










  • Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
    – HectorJ
    1 hour ago






  • 1




    @HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
    – François Andrieux
    1 hour ago


















10














The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.



Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.






share|improve this answer























  • Since c++11 no? unless I’ve lost my marbles.
    – Bathsheba
    1 hour ago










  • Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
    – HectorJ
    1 hour ago






  • 1




    @HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
    – François Andrieux
    1 hour ago
















10












10








10






The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.



Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.






share|improve this answer














The throw() in void f() throw() is a dynamic exception specification and is deprecated since c++11. It's was supposed to be used to list the exceptions a function could throw. An empty specification (throw()) means there are no exceptions that your function could throw. Trying to throw an exception from such a function calls std::unexpected which, by default, terminates.



Since c++11 the preferred way of specifying that a function cannot throw is to use noexcept. For example void f() noexcept.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









François Andrieux

15.3k32547




15.3k32547












  • Since c++11 no? unless I’ve lost my marbles.
    – Bathsheba
    1 hour ago










  • Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
    – HectorJ
    1 hour ago






  • 1




    @HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
    – François Andrieux
    1 hour ago




















  • Since c++11 no? unless I’ve lost my marbles.
    – Bathsheba
    1 hour ago










  • Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
    – HectorJ
    1 hour ago






  • 1




    @HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
    – François Andrieux
    1 hour ago


















Since c++11 no? unless I’ve lost my marbles.
– Bathsheba
1 hour ago




Since c++11 no? unless I’ve lost my marbles.
– Bathsheba
1 hour ago












Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
– HectorJ
1 hour ago




Thank you! I had an incorrect understanding. It looks like noexcept(false) will provide the same functionality and looks to not have been deprecated
– HectorJ
1 hour ago




1




1




@HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
– François Andrieux
1 hour ago






@HectorJ noexcept(false) is the default for most functions and usually does not need to be provided. I believe destructors are noexcept(true) by default and as far as I know that would be the only time noexcept(false) has meaning (though that's not recommended).
– François Andrieux
1 hour ago




















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