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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     

Group: comp.std.c++ · Group Profile · Search for Sashan in comp.std.c++
Author: Gabriel Dos Reis
Date: Jul 17, 2006 15:16

On Mon, 17 Jul 2006, Gennaro Prota wrote: | On Mon, 17 Jul 2006 00:58:47 CST, Gabriel Dos Reis <gdr@cs.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
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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     

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Author: Gabriel Dos Reis
Date: Jul 17, 2006 13:43

On Mon, 17 Jul 2006 00:58:47 CST, Gabriel Dos Reis <gdr@cs.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
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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     

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Author: James Dennett
Date: Jul 17, 2006 12:35

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
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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     

Group: comp.std.c++ · Group Profile · Search for Sashan in comp.std.c++
Author: Gennaro Prota
Date: Jul 17, 2006 10:44

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
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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     

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Author: Gabriel Dos Reis
Date: Jul 16, 2006 23:58

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Author: Gabriel Dos Reis
Date: Jul 16, 2006 22:59

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Author: Gabriel Dos Reis
Date: Jul 16, 2006 19:33

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Group: comp.std.c++ · Group Profile · Search for Sashan in comp.std.c++
Author: Peter C. Chapin
Date: Jul 15, 2006 06:56

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Group: comp.std.c++ · Group Profile · Search for Sashan in comp.std.c++
Author: James Dennett
Date: Jul 15, 2006 05:42

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