On Mon, 20 Jan 2003 15:04:30 -0800, "Eric Niebler" <[EMAIL PROTECTED]> wrote:
>It seems to me that the current implementation of is_convertible will >conclude that int* is convertible to int[10]. That's because when a >function parameter is of array type, it is treated just like a pointer. But >according to 4/3 of the standard: > >"An expression e can be _implicitly converted_ to a type T if an only if the >declaration "T t=e;" is well-formed for some invented variable t (8.5)." > >By this definition, int* is not implicitly convertible to int[10]. So it >seems is_convertible is broken in this regard. IMO, is_convertible<A,B> >should always return false when B is an array type. I don't want to be involved in another war here, also because I'm totally ignorant about boost's implementation of is_convertible and because I don't know what is its purpose (i.e.: I haven't read the documentation and I can just have an idea from its name). I want simply to point out that saying a type is or isn't convertible to another has a very vague meaning. Note that the definition you quote from the standard concerns conversion of an *expression*, not of a type. Also it just concerns *implicit* conversion. So, off the top of my head, "is convertible" means nothing to me (though I can imagine how it is implemented). NOTE: I'm not saying there's an error in boost. It's likely that the documentation explains what its meaning is, but I don't have time to check now. Note anyway that e.g. (1-1) is an expression of type int and that it is implicitly convertible to void*, to void (*) int, or to char (A::*)(int) but I doubt that is_convertible would yield true on the corresponding types. The exact meaning of is_convertible is likely to be understood only if you look at the implementation I think, unless there's a very good, and technical, explanation in the docs. Genny. _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
