C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  What is the meaning of the no throw?          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Julius Guest Posted: Wed Jan 12, 2005 9:13 pm    Post subject: What is the meaning of the no throw? Hello people, I am just curious of meaning of throw()? Like in function signature: void foo(void) throw(); I understand that it means that this certain function will never throw any exception. Right? But why can I in the body of the function explicetely throw? At least VS7 allow this I have only warning, no more. For example: void foo (void ) throw() { throw true; } Thanks in advance. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Victor Bazarov Guest Posted: Thu Jan 13, 2005 12:08 pm    Post subject: Re: What is the meaning of the no throw? "Julius" <yuliy (AT) gmx (DOT) de> wrote... Quote: I am just curious of meaning of throw()? Like in function signature: void foo(void) throw(); It's a promise that the function does not throw any C++ exceptions. Quote: I understand that it means that this certain function will never throw any exception. Right? Right. Quote: But why can I in the body of the function explicetely throw? At least VS7 allow this I have only warning, no more. For example: void foo (void ) throw() { throw true; } It is expressly allowed to throw exceptions that the function promises not to throw (in this case, any). The behaviour is to call 'unexpected()' if the exception is thrown that is not declared in the function's exception specification (see 15.4) Victor [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top ntrifunovic@hotmail.com Guest Posted: Thu Jan 13, 2005 12:08 pm    Post subject: Re: What is the meaning of the no throw? Look here for an excellent explanation: http://www.gotw.ca/publications/mill22.htm [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Markus Moll Guest Posted: Thu Jan 13, 2005 12:12 pm    Post subject: Re: What is the meaning of the no throw? Hi Julius wrote: Quote: I understand that it means that this certain function will never throw any exception. Right? Yes. But see below... Quote: But why can I in the body of the function explicetely throw? At least VS7 allow this I have only warning, no more. There's no problem with that. Consider the more general case of non-empty throw-specifiers. Whenever an exception would leave a block with a throws specifier disallowing the type of that exception, the "std::unexpected()" function is called. You can set your own handler via "set_unexpected(unexpected_handler h)". The standard unexpected handler tries to throw a std::bad_exception object, if this is also disallowed it calls "terminate()". Try the following: #include <iostream> void f() throw (int) { throw 1.0; } void whoa() { throw 42; } int main() { set_unexpected(whoa); try { f(); } catch(int i) { std::cout << i << std::endl; } } Markus [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top L.Suresh Guest Posted: Thu Jan 13, 2005 4:38 pm    Post subject: Re: What is the meaning of the no throw? Right, it is a strong guarantee that it wont throw anything. If it violates this guarantee then std::unexpected() is called which by default calls std::terminate() which in turn calls abort(). The example you have given is equivalent to ... void foo() throw() { try { throw true; } catch (...) { std::unexpected(); } } --lsu [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top L.Suresh Guest Posted: Thu Jan 13, 2005 4:39 pm    Post subject: Re: What is the meaning of the no throw? Right, it is a strong guarantee that it wont throw anything. If it violates this guarantee then std::unexpected() is called which by default calls std::terminate() which in turn calls abort(). The example you have given is equivalent to ... void foo() throw() { try { throw true; } catch (...) { std::unexpected(); } } --lsu [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Richter Guest Posted: Thu Jan 13, 2005 5:10 pm    Post subject: Re: What is the meaning of the no throw? Hi, Quote: I am just curious of meaning of throw()? Like in function signature: void foo(void) throw(); "Dear compiler, I promise not to throw in this routine and explicitly allow you to take advantage of this". Thus, usually not much better than a comment, IMHO, except that it possibly might (or might not) help the compiler a bit. Quote: I understand that it means that this certain function will never throw any exception. Right? It means that you hand out a contract to the compiler that this function never throws. If it does, anything could happen. Quote: But why can I in the body of the function explicetely throw? Because whether a function is able to throw or not is out of the control of the compiler anyhow. You could call other functions that might or might not throw, possibly in external libraries. Or, what about the following: Will this function satisfy the contract of not throwing ? #include <cmath> void foo(void) throw() { if (sqrt(9.0) != 3.0) throw true; } or, if that sounds too obvious, what about this one: #include <cmath> void foo(void) throw() { if (sin(3.14159326) != 0.0) throw true; } IMHO, throw() is not a very wise construction. Avoid it. So long, Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top GianGuz Guest Posted: Thu Jan 13, 2005 5:18 pm    Post subject: Re: What is the meaning of the no throw? Julius wrote: Quote: Hello people, I am just curious of meaning of throw()? Like in function signature: void foo(void) throw(); I understand that it means that this certain function will never throw any exception. Right? But why can I in the body of the function explicetely throw? At least VS7 allow this I have only warning, no more. For example: void foo (void ) throw() { throw true; } Thanks in advance. Throw list specification is always a bad idea. When a function is marked with a throw specifier some upsetting things happened: 1) The throw(x,y,...,z) signature becomes part of the function type - so f() throw() is different from f() 2) The compiler is 'suggested' to optimize code only for exceptions hinted by the throw list and if it is able to infer that really no exception will be throw (i.e. f() throw() ) it will remove the exception handling into that scope. 3) 'An "unexpected exception" is called when a function throws an exception it promised not to throw, in violation of its exception specification. The unexpected exception may be replaced by a bad_exception object during the stack unwind' [ Schmidt ] As you pointed out, nobody prevents you to throw any exceptions into your functions even if it is forbidden by the exception specification (problem 3). By the way the compiler is able to remove useless try catcth block without exceptions specification (problem 2). So use that language feature only when you have to interface to code that is using it and you can't avoid type mismatching (problem 1). Gianguglielmo [excess quoting deleted --mod] [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Victor Bazarov Guest Posted: Thu Jan 13, 2005 9:59 pm    Post subject: Re: What is the meaning of the no throw? L.Suresh wrote: Quote: Right, it is a strong guarantee that it wont throw anything. If it violates this guarantee then std::unexpected() is called which by default calls std::terminate() which in turn calls abort(). 'unexpected' is allowed to re-throw instead of calling 'terminate'. V [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Markus Moll Guest Posted: Sat Jan 15, 2005 3:33 am    Post subject: Re: What is the meaning of the no throw? Hi Thomas Richter wrote: Quote: It means that you hand out a contract to the compiler that this function never throws. If it does, anything could happen. No. The behaviour is defined. Quote: IMHO, throw() is not a very wise construction. Avoid it. IMHO, it's the right thing to do if you a) do not want that function to throw anything b) want to let the reader know that it's useless to try to catch any exceptions here c) want to force derivations not to throw anything either For example, this gives you the opportunity to have simple exception-safe code, eg in void f(SomeClass*) throw(); SomeClass *p = new SomeClass(); f(p); delete p; (Yes, there _is_ std::auto_ptr... and yes, the example is kind of pointless...) Worst thing that could happen is the program terminates. Markus [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Dave Moore Guest Posted: Sat Jan 15, 2005 3:42 am    Post subject: Re: What is the meaning of the no throw? "Thomas Richter" <thor (AT) cleopatra (DOT) math.tu-berlin.de> wrote Quote: Hi, I am just curious of meaning of throw()? Like in function signature: void foo(void) throw(); "Dear compiler, I promise not to throw in this routine and explicitly allow you to take advantage of this". No, the no-throw guarantee says that, "This function is *always* expected to return normally. If it does not, it means that something outside the subset of events this software is designed to handle has happened." Quote: Thus, usually not much better than a comment, IMHO, except that it possibly might (or might not) help the compiler a bit. It is much better than a comment, because it activates a fail-safe mechanism, namely calling std::unexpected(), immediately at the point that the unexpected event happened. This may be more appropriate than allowing the unexpected exception to propagate up the stack. Basically your position (which seems to be shared by many other respondants in this thread), is tantamount to saying "All exceptions are unrecoverable errors, so don't worry about classifying them (except for returning error messages perhaps)." The no-throw guarantee, and exception specifications in general, provide a built-in way for a program to separate "Ok, that is exceptional behavior I can (maybe) deal with." from "Holy crap, what the heck just happened!!". To replicate such a mechanism without it, you would have to call std::unexpected() manually from inside a catch(...) block after every function call that meets the criterion "should never fail to return". Quote: I understand that it means that this certain function will never throw any exception. Right? It means that you hand out a contract to the compiler that this function never throws. If it does, anything could happen. Wrong. The behavior is quite specifically defined: std::unexpected() is called, which by default in turn calls std::terminate(). However the user can override this behavior by specifying an alternate handler using set_unexpected(). [snip] Quote: IMHO, throw() is not a very wise construction. Avoid it. Well, the developers of the standard library would disagree I guess, because they used no-throw exception specifications quite frequently. OTOH, I have never needed to use them myself, but then I have never had to deal with anything more critical than handling errors in user-input, or floating point exceptions. Perhaps better advice would be, "make sure a throw() specification is really warranted, and that you understand the ramifications", which boils down to "think carefully about your design". I guess no one will argue with that 8*). Dave Moore [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top L.Suresh Guest Posted: Sat Jan 15, 2005 4:10 am    Post subject: Re: What is the meaning of the no throw? Quote: 'unexpected' is allowed to re-throw instead of calling 'terminate'. 15.5.1, terminate() is called when "- when the implementation's default unexpected_handler is called" The default behaviour is to call std::terminate(). Of course, you can write your own unexpected_handler that does things differently. --lsu [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top natkhatbandar@yahoo.com Guest Posted: Sun Jan 16, 2005 4:14 am    Post subject: Re: What is the meaning of the no throw? Just FYI: VC++ 6.0 does not call unexpected() at all !!! The function that calls foo() will be able to catch the exception of type bool. The documentation admits that unexpected() is never called by the exception-handling run-time library. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Richter Guest Posted: Tue Jan 18, 2005 11:13 pm    Post subject: Re: What is the meaning of the no throw? Hi Dave, Quote: "Dear compiler, I promise not to throw in this routine and explicitly allow you to take advantage of this". No, the no-throw guarantee says that, "This function is *always* expected to return normally. If it does not, it means that something outside the subset of events this software is designed to handle has happened." Thus, usually not much better than a comment, IMHO, except that it possibly might (or might not) help the compiler a bit. It is much better than a comment, because it activates a fail-safe mechanism, namely calling std::unexpected(), immediately at the point that the unexpected event happened. This may be more appropriate than allowing the unexpected exception to propagate up the stack. Thanks, I stand corrected. Quote: Basically your position (which seems to be shared by many other respondants in this thread), is tantamount to saying "All exceptions are unrecoverable errors, so don't worry about classifying them (except for returning error messages perhaps)." Not quite. My position is that somewhere in the main, you need to put a catch() and print there that you caught an exception that shouldn't have propagated up to that level. (-; Quote: The no-throw guarantee, and exception specifications in general, provide a built-in way for a program to separate "Ok, that is exceptional behavior I can (maybe) deal with." from "Holy crap, what the heck just happened!!". Strange point of view... If I can deal with the exception, I put a catch(...), and no throw(). If the function throws despite a throw(), it breaks its contract to the caller. If I would want to get a handler called, I could have done that myself as an error reporting mechanism in first place. Quote: To replicate such a mechanism without it, you would have to call std::unexpected() manually from inside a catch(...) block after every function call that meets the criterion "should never fail to return". Yes, exactly. Except that a function that should never fail should never call std::unexpected either. It is just a different mechanism of reporting an exceptional condition for something the function promised not to cause an exception in first place. That is, I believe there's a basic design flaw in std::unexpected(): It is a way of bypassing the contract of the function. Quote: It means that you hand out a contract to the compiler that this function never throws. If it does, anything could happen. Wrong. The behavior is quite specifically defined: std::unexpected() is called, which by default in turn calls std::terminate(). However the user can override this behavior by specifying an alternate handler using set_unexpected(). Thanks for the clarification. Quote: IMHO, throw() is not a very wise construction. Avoid it. Well, the developers of the standard library would disagree I guess, because they used no-throw exception specifications quite frequently. OTOH, I have never needed to use them myself, but then I have never had to deal with anything more critical than handling errors in user-input, or floating point exceptions. Perhaps better advice would be, "make sure a throw() specification is really warranted, and that you understand the ramifications", which boils down to "think carefully about your design". I guess no one will argue with that 8*). True enough. (-: So long, Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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