C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Why won't the compiler tell          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Mark Rance Guest Posted: Wed Apr 28, 2004 7:28 pm    Post subject: Why won't the compiler tell Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? The only way this works is if I have declared a variable of type foo...yet the compiler knows quite well what the answer to what I am asking. This just seems like something the compilers should be able to handle. -Mark [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Ralf Guest Posted: Thu Apr 29, 2004 11:35 am    Post subject: Re: Why won't the compiler tell The sizeof operator needs either a type or a variable. What you gave it is nothing of the two possibilities. www.oop-trainer.de [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Niels Dybdahl Guest Posted: Thu Apr 29, 2004 9:24 pm    Post subject: Re: Why won't the compiler tell Quote: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? Some compilers can do that (MS VC). Niels Dybdahl [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top llewelly Guest Posted: Thu Apr 29, 2004 9:37 pm    Post subject: Re: Why won't the compiler tell "Mark Rance" <mrr (AT) pcisys (DOT) network> writes: Quote: Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? [snip] The operand of sizeof must be an expression. See 5.3.3 . 'foo.b' is not a well-formed expression, since 'foo' is a type, not an object. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Glen Low Guest Posted: Fri Apr 30, 2004 11:14 am    Post subject: Re: Why won't the compiler tell Quote: Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? If you had done struct foo { char a [10]; char b [5]; char c [3]; }; then you can simply do size_t x = sizeof (foo::b); Cheers, Glen Low, Pixelglow Software www.pixelglow.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Paul D. DeRocco Guest Posted: Fri Apr 30, 2004 11:21 am    Post subject: Re: Why won't the compiler tell Quote: "Mark Rance" <mrr (AT) pcisys (DOT) network> wrote Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? The only way this works is if I have declared a variable of type foo...yet the compiler knows quite well what the answer to what I am asking. This just seems like something the compilers should be able to handle. I would think that the correct syntax would be sizeof(foo::b), but that's illegal, too. However, you can kluge it with: size_t x = sizeof(((foo*)0)->b); -- Ciao, Paul D. DeRocco Paul mailto:pderocco (AT) ix (DOT) netcom.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Mark Rance Guest Posted: Fri Apr 30, 2004 11:35 am    Post subject: Re: Why won't the compiler tell "Niels Dybdahl" <ndy (AT) fjern (DOT) detteesko-graphics.com> wrote Quote: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? Some compilers can do that (MS VC). Not true. I am using the VC7 compiler and it does not work. This does, however, size_t x = (((foo *) 0)->b); -Mark [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Mark Rance Guest Posted: Fri Apr 30, 2004 11:35 am    Post subject: Re: Why won't the compiler tell "llewelly" <llewelly.at (AT) xmission (DOT) dot.com> wrote Quote: "Mark Rance" <mrr (AT) pcisys (DOT) network> writes: Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? [snip] The operand of sizeof must be an expression. See 5.3.3 . 'foo.b' is not a well-formed expression, since 'foo' is a type, not an object. I know, but it just seemed odd since the compiler DOES know the answer...I thought I should be able to get away with it. I posted the work-around I used in an earlier post to this thread, if interested. -Mark [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Francis Glassborow Guest Posted: Sat May 01, 2004 3:09 am    Post subject: Re: Why won't the compiler tell In message <aBnkc.41$fh4.11 (AT) newsread1 (DOT) news.pas.earthlink.net>, Paul D. DeRocco <pderocco (AT) ix (DOT) netcom.com> writes Quote: I would think that the correct syntax would be sizeof(foo::b), but that's illegal, too. However, you can kluge it with: size_t x = sizeof(((foo*)0)->b); I think that is the one case where (foo *)0 is OK because sizeof does not evaluate the expression. But what puzzles me is why you would ever want to do this because you already know what b's type is and so can simply ask for the sizeof the type. -- Francis Glassborow ACCU Author of 'You Can Do It!' see http://www.spellen.org/youcandoit For project ideas and contributions: http://www.spellen.org/youcandoit/projects [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze@gabi-soft.fr Guest Posted: Mon May 03, 2004 2:51 pm    Post subject: Re: Why won't the compiler tell "Mark Rance" <mrr (AT) pcisys (DOT) network> wrote Quote: "Niels Dybdahl" <ndy (AT) fjern (DOT) detteesko-graphics.com> wrote in message news:4090b42b$0$164$edfadb0f (AT) dtext02 (DOT) news.tele.dk... typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? Some compilers can do that (MS VC). Not true. I am using the VC7 compiler and it does not work. This does, however, size_t x = (((foo *) 0)->b); You mean: size_t x = sizeof( ((foo*)0)->b ) ; don't you. (What you have written should not compile.) -- James Kanze GABI Software mailto:kanze (AT) gabi-soft (DOT) fr Conseils en informatique orientée objet/ http://www.gabi-soft.fr Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze@gabi-soft.fr Guest Posted: Mon May 03, 2004 2:51 pm    Post subject: Re: Why won't the compiler tell Francis Glassborow <francis (AT) robinton (DOT) demon.co.uk> wrote Quote: In message <aBnkc.41$fh4.11 (AT) newsread1 (DOT) news.pas.earthlink.net>, Paul D. DeRocco <pderocco (AT) ix (DOT) netcom.com> writes I would think that the correct syntax would be sizeof(foo::b), but that's illegal, too. However, you can kluge it with: size_t x = sizeof(((foo*)0)->b); I think that is the one case where (foo *)0 is OK because sizeof does not evaluate the expression. But what puzzles me is why you would ever want to do this because you already know what b's type is and so can simply ask for the sizeof the type. In the case in question, b's type was char[5], or something like that. You don't want to have the magic number appearing in more than one place. -- James Kanze GABI Software mailto:kanze (AT) gabi-soft (DOT) fr Conseils en informatique orientée objet/ http://www.gabi-soft.fr Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Michiel Salters Guest Posted: Tue May 04, 2004 10:37 am    Post subject: Re: Why won't the compiler tell "Mark Rance" <mrr (AT) pcisys (DOT) network> wrote Quote: Consider this: typedef struct { char a[10]; char b[5]; char c[3]; } foo; Why can I not execute a statement like: size_t x = sizeof(foo.b); ?? The only way this works is if I have declared a variable of type foo...yet the compiler knows quite well what the answer to what I am asking. This just seems like something the compilers should be able to handle. In addition to what Francis noted, in C++ we'd probably write struct foo { typedef char a_type[10]; typedef char b_type[5]; typedef char c_type[3]; a_type a; b_type b; c_type c; }; size_t x = sizeof( foo::a_type ); Regards, Michiel Salters [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Balog Pal Guest Posted: Sun May 09, 2004 5:59 pm    Post subject: Re: Why won't the compiler tell "Francis Glassborow" <francis (AT) robinton (DOT) demon.co.uk> wrote Quote: size_t x = sizeof(((foo*)0)->b); I think that is the one case where (foo *)0 is OK because sizeof does not evaluate the expression. But what puzzles me is why you would ever want to do this because you already know what b's type is and so can simply ask for the sizeof the type. That would introduce unwanted redundancy. The whole thing is a good example for why we miss typeof() so much. In a solid program I could change type of foo::b at that only location in the class definition, and everything keep working. The suggestion to use typedefs from the other post is pretty ugly if that typedef doeesn't make sense in itself, what is a frequent case. Paul [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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