C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  rationale behind typename          C++Talk.NET Forum Index -> C++ language, library and standards View previous topic :: View next topic   Author Message sashan Guest Posted: Sat Jul 15, 2006 7:19 am    Post subject: rationale behind typename Hi I'm wondering why the std specifies that typename be used in cases like this: typename Foo<T>::bar b2; When bar is a type as opposed to a variable? Why can't the compiler make that decision? --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Peter C. Chapin Guest Posted: Sat Jul 15, 2006 1:56 pm    Post subject: Re: rationale behind typename sashan wrote: Quote: I'm wondering why the std specifies that typename be used in caseslike this: typename Foo<T>::bar b2; When bar is a type as opposed to a variable? Why can't the compiler make that decision? The typename is only needed inside the definition of a template. When the compiler processes the template it doesn't know what type is represented by T (that happens later during instantiation). Consequently it can't tell of bar is a type or a an object. Consider: template< typename T > struct Foo { typedef int bar; }; // Specialized version for int. template<> struct Foo<int> { static int bar; }; In the general template the member bar is a type. However, there is a specialization for which the member bar is an object. Now consider Foo<T>::bar. If T is int this refers to an object, otherwise it refers to a type. When the compiler processes the template that contains Foo<T>::bar it doesn't know (yet) what it is dealing with and so you must explicitly tell it. If you include typename, Foo<T>::bar is a type, otherwise it is an object. Be aware that in general C++ (and C) can't be parsed without knowing what identifiers are the names of types. This business with typename wasn't necessary in old C++ because in old C++ template bodies were not fully parsed as they were read. They were only parsed at instantiation time when the nature of the template arguments was known. In modern C++ we have two phase lookup. Non-dependent names in the template bodies are bound to declarations at the time the template itself is seen. To do this, the template must be fully parsed before the arguments are known and this leads to the problem that typename solves. Hope this helps! Peter --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top James Dennett Guest Posted: Sat Jul 15, 2006 5:42 pm    Post subject: Re: rationale behind typename sashan wrote: Quote: Hi I'm wondering why the std specifies that typename be used in cases like this: typename Foo<T>::bar b2; When bar is a type as opposed to a variable? Why can't the compiler make that decision? Because, assuming T is a dependent type (i.e., a type depending on a template parameter), it's unknowable at the time the template is parsed. There may be some T for which Foo<T>::bar is a type, and others for which it is an integral constant expression (and yet others for which it does not exist, though that's not so relevant here). In order to allow templates to be checked for syntax, the compiler must know for each identifier in the template whether that identifier refers to a type and whether it refers to a template. This is why the keywords typename and template are sometimes needed in generic code, when they are not required in non-generic code. -- James --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gabriel Dos Reis Guest Posted: Mon Jul 17, 2006 6:58 am    Post subject: Re: rationale behind typename On Sat, 15 Jul 2006, sashan wrote: | Hi | | I'm wondering why the std specifies that typename be used in cases like | this: | | typename Foo<T>::bar b2; unfortunate limitation of parsing technology that shrines in the language specification. | When bar is a type as opposed to a variable? Why can't the compiler | make that decision? Consider the following slightly modified case Foo<T>::bar * foo; declaration of pointer "foo" or multiplication? -- Gabriel Dos Reis gdr (AT) cs (DOT) tamu.edu Texas A&M University -- Department of Computer Science 301, Bright Building -- College Station, TX 77843-3112 --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gabriel Dos Reis Guest Posted: Mon Jul 17, 2006 7:33 am    Post subject: Re: rationale behind typename On Sat, 15 Jul 2006, Peter C. Chapin wrote: | sashan wrote: | | > I'm wondering why the std specifies that typename be used in caseslike | > this: | > | > typename Foo<T>::bar b2; | > | > When bar is a type as opposed to a variable? Why can't the compiler | > make that decision? | | The typename is only needed inside the definition of a template. When | the compiler processes the template it doesn't know what type is | represented by T (that happens later during instantiation). Consequently | it can't tell of bar is a type or a an object. Consider: | | template< typename T > | struct Foo { | typedef int bar; | }; | | // Specialized version for int. | template<> | struct Foo<int> { | static int bar; | }; Yes, but that does not explain why "typename" is needed at the *parsing time*. The declaration typename Foo<T>::bar foo; is invalid for instantiation of Foo with T=int -- assuming your specializations. Now, assume the committee decided that Foo<T>::bar -- without the preceding "typename" -- shall be interpreted as a typename. The above program with behave the same as if the "typename" was there. However, not requiring the "typename" means that there is no way to say F<T>::B * f; is actually a multiplication, or that F<T>::f(x); is a function call -- remember, C and thereofre C++, allows redundant parenthesis in declarations. -- Gabriel Dos Reis gdr (AT) cs (DOT) tamu.edu Texas A&M University -- Department of Computer Science 301, Bright Building -- College Station, TX 77843-3112 --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gabriel Dos Reis Guest Posted: Mon Jul 17, 2006 9:10 am    Post subject: Re: rationale behind typename On Sat, 15 Jul 2006, James Dennett wrote: | sashan wrote: | > Hi | > | > I'm wondering why the std specifies that typename be used in cases like | > this: | > | > typename Foo<T>::bar b2; | > | > When bar is a type as opposed to a variable? Why can't the compiler | > make that decision? | | Because, assuming T is a dependent type (i.e., a type depending | on a template parameter), it's unknowable at the time the template | is parsed. There may be some T for which Foo<T>::bar is a type, | and others for which it is an integral constant expression (and | yet others for which it does not exist, though that's not so | relevant here). but even in those disparate cases, using "typename" does nto make the resulting program valid. Consider Foo<T>::bar * foo; or Foo<T>::bar(x); Irrespective of whether the instantiations of Foo<T> may behave as you said, the parser --before going into instantiation considerations-- still needs clues for deciding whether to parse them as declarations or expressions-statements. -- Gabriel Dos Reis gdr (AT) cs (DOT) tamu.edu Texas A&M University -- Department of Computer Science 301, Bright Building -- College Station, TX 77843-3112 --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gabriel Dos Reis Guest Posted: Mon Jul 17, 2006 10:16 pm    Post subject: Re: rationale behind typename On Mon, 17 Jul 2006, James Dennett wrote: | Gabriel Dos Reis wrote: | > On Sat, 15 Jul 2006, James Dennett wrote: | > | > | sashan wrote: | > | > Hi | > | > | > | > I'm wondering why the std specifies that typename be used in cases like | > | > this: | > | > | > | > typename Foo<T>::bar b2; | > | > | > | > When bar is a type as opposed to a variable? Why can't the compiler | > | > make that decision? | > | | > | Because, assuming T is a dependent type (i.e., a type depending | > | on a template parameter), it's unknowable at the time the template | > | is parsed. There may be some T for which Foo<T>::bar is a type, | > | and others for which it is an integral constant expression (and | > | yet others for which it does not exist, though that's not so | > | relevant here). | > | > but even in those disparate cases, using "typename" does nto make the | > resulting program valid. | | I'm afraid I can't understand that; typename is in the language | precisely because it makes the program valid, so long as the | template is used only for suitable types, and because it allows | the template to be parsed without additional knowledge of the | types over which it will be instantiated. My message was saying that putting "typename" in front of the code does not make it legal if you instantiate Foo with T=int. That you can't dispute. I was clarifying that "typename" has no bearing to instantiation as several messages tried to imply. It is a pure parsing problem, not instantiation problem. | You snipped the paragraph I wrote explaining that in more | detail: | | In order to allow templates to be checked for syntax, | the compiler must know for each identifier in the | template whether that identifier refers to a type and | whether it refers to a template. This is why the | keywords typename and template are sometimes needed | in generic code, when they are not required in | non-generic code. This is alsmot correct; but the earlier instantiation stuff I find confusing In fact, it is not really checking for syntax as opposed to letting the compiler proceed in the parsing. The compiler checks almost nothing in Foo<T>::bar(x); or typename Foo<T>::bar(x); It trusts the programmer. | > Consider | > | > Foo<T>::bar * foo; | > | > or | > | > Foo<T>::bar(x); | > | > Irrespective of whether the instantiations of Foo<T> may behave as you | > said, the parser --before going into instantiation considerations-- | > still needs clues for deciding whether to parse them as declarations | > or expressions-statements. | | Yes, which is why typename is used to distinguish which case | your template intends to support. Without typename, bar is | not a type name in the above, so we have multiplication or | a function call (expression statements), but with typename | they become declarations (with redundant () in the latter). I know all that. The question was "what if it wasn't required" :-) -- Gabriel Dos Reis gdr (AT) cs (DOT) tamu.edu Texas A&M University -- Department of Computer Science 301, Bright Building -- College Station, TX 77843-3112 --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gennaro Prota Guest Posted: Mon Jul 17, 2006 10:44 pm    Post subject: Re: rationale behind typename On Mon, 17 Jul 2006 00:58:47 CST, Gabriel Dos Reis <gdr (AT) cs (DOT) tamu.edu> wrote: Quote: On Sat, 15 Jul 2006, sashan wrote: | Hi | | I'm wondering why the std specifies that typename be used in cases like | this: | | typename Foo<T>::bar b2; unfortunate limitation of parsing technology that shrines in the language specification. So it might be made optional in such cases, in the future? :) -- Gennaro Prota --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top James Dennett Guest Posted: Tue Jul 18, 2006 12:35 am    Post subject: Re: rationale behind typename Gabriel Dos Reis wrote: Quote: On Sat, 15 Jul 2006, James Dennett wrote: | sashan wrote: | > Hi | | > I'm wondering why the std specifies that typename be used in cases like | > this: | | > typename Foo<T>::bar b2; | | > When bar is a type as opposed to a variable? Why can't the compiler | > make that decision? | | Because, assuming T is a dependent type (i.e., a type depending | on a template parameter), it's unknowable at the time the template | is parsed. There may be some T for which Foo<T>::bar is a type, | and others for which it is an integral constant expression (and | yet others for which it does not exist, though that's not so | relevant here). but even in those disparate cases, using "typename" does nto make the resulting program valid. I'm afraid I can't understand that; typename is in the language precisely because it makes the program valid, so long as the template is used only for suitable types, and because it allows the template to be parsed without additional knowledge of the types over which it will be instantiated. You snipped the paragraph I wrote explaining that in more detail: In order to allow templates to be checked for syntax, the compiler must know for each identifier in the template whether that identifier refers to a type and whether it refers to a template. This is why the keywords typename and template are sometimes needed in generic code, when they are not required in non-generic code. Quote: Consider Foo<T>::bar * foo; or Foo<T>::bar(x); Irrespective of whether the instantiations of Foo<T> may behave as you said, the parser --before going into instantiation considerations-- still needs clues for deciding whether to parse them as declarations or expressions-statements. Yes, which is why typename is used to distinguish which case your template intends to support. Without typename, bar is not a type name in the above, so we have multiplication or a function call (expression statements), but with typename they become declarations (with redundant () in the latter). -- James --- [ comp.std.c++ is moderated. To submit articles, try just posting with ] [ your news-reader. If that fails, use mailto:std-c++@ncar.ucar.edu ] [ --- Please see the FAQ before posting. --- ] [ FAQ: http://www.comeaucomputing.com/csc/faq.html ] Back to top Gabriel Dos Reis Guest Posted: Tue Jul 18, 2006 1:43 am    Post subject: Re: rationale behind typename On Mon, 17 Jul 2006, Gennaro Prota wrote: | On Mon, 17 Jul 2006 00:58:47 CST, Gabriel Dos Reis <gdr (AT) cs (DOT) tamu.edu> | wrote: | | >On Sat, 15 Jul 2006, sashan wrote: | > | >| Hi | >| | >| I'm wondering why the std specifies that typename be used in cases like | >| this: | >| | >| typename Foo<T>::bar b2; | > | >unfortunate limitation of parsing technology that shrines in the | >language specification. | | So it might be made optional in such cases, in the future? :) I see your smiley; however I must add that I don't see how making it optional will decrease the confusion around "typename". -- Gabriel Dos Reis gdr (AT) cs (DOT) tamu.edu Texas A&M University -- Department of Computer Science 301, Bright Building -- College Station, TX 77843-3112 --- [ comp.std.c++ is moderated. 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