C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Foo(300) = Foo(500); Why does a compiler compile that?          C++Talk.NET Forum Index -> C++ language (comp.lang.c++) View previous topic :: View next topic   Author Message Alex Vinokur Guest Posted: Sun Jan 30, 2005 9:48 am    Post subject: Foo(300) = Foo(500); Why does a compiler compile that? Compiler GNU gpp.exe (GCC) 3.4.1 Foo(300) = Foo(500); // Foo(300) is const. Why does a compiler compile that? ------ foo.cpp ------ struct Foo { explicit Foo(int) {} Foo& operator= (const Foo&) { return *this; } }; int main() { Foo var1(100); const Foo var2(200); var1 = Foo(500); Foo(300) = Foo(500); // Foo(300) is const. Why does a compiler compile that? // var2 = Foo(500); // A compiler doesn't compile that return 0; } --------------------- -- Alex Vinokur email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn Back to top Ivan Vecerina Guest Posted: Sun Jan 30, 2005 9:57 am    Post subject: Re: Foo(300) = Foo(500); Why does a compiler compile that? "Alex Vinokur" <alexvn (AT) big-foot (DOT) com> wrote Quote: Compiler GNU gpp.exe (GCC) 3.4.1 Foo(300) = Foo(500); // Foo(300) is const. Why does a compiler compile that? Foo(300) is not const - it is a temporary. It is true that temporaries can be bound to a const reference only (not to a non-const ref). However, it is legal to call any member function on a temporary. And the line above is equivalent to: Foo(300).operator=( Foo(500) ); This is indeed a case where user-defined classes do not behave like built-in types... Cheers, Ivan -- http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form Back to top Alex Vinokur Guest Posted: Sun Jan 30, 2005 10:09 am    Post subject: Re: Foo(300) = Foo(500); Why does a compiler compile that? "Ivan Vecerina" <INVALID_use_webform_instead (AT) vecerina (DOT) com> wrote Quote: "Alex Vinokur" <alexvn (AT) big-foot (DOT) com> wrote in message news:363oriF4scb34U1 (AT) individual (DOT) net... Compiler GNU gpp.exe (GCC) 3.4.1 Foo(300) = Foo(500); // Foo(300) is const. Why does a compiler compile that? Foo(300) is not const - it is a temporary. ---------------------------------------- Quote: It is true that temporaries can be bound to a const reference only (not to a non-const ref). If 'Foo(300) = Foo(500);' is legal, why can temporaries be bound to a const reference only (not to a non-const ref)? --------------------------------------- Quote: However, it is legal to call any member function on a temporary. And the line above is equivalent to: Foo(300).operator=( Foo(500) ); This is indeed a case where user-defined classes do not behave like built-in types... [snip] -- Alex Vinokur email: alex DOT vinokur AT gmail DOT com http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn Back to top Ivan Vecerina Guest Posted: Sun Jan 30, 2005 10:54 am    Post subject: Re: Foo(300) = Foo(500); Why does a compiler compile that? "Alex Vinokur" <alexvn (AT) big-foot (DOT) com> wrote Quote: "Ivan Vecerina" <INVALID_use_webform_instead (AT) vecerina (DOT) com> wrote in message news:ctib29$n0u$1 (AT) news (DOT) hispeed.ch... Foo(300) = Foo(500); // Foo(300) is const. Why does a compiler compile that? Foo(300) is not const - it is a temporary. ---------------------------------------- It is true that temporaries can be bound to a const reference only (not to a non-const ref). If 'Foo(300) = Foo(500);' is legal, why can temporaries be bound to a const reference only (not to a non-const ref)? Probably to avoid programming errors such as: copySomeMembers( myTargetVariable, getSomeTemporary() ); //woops: actually, the functions takes 1st source, 2nd destination: // copySomeMembers( Foo const& src, Foo& dst ); Because temps can only be bound to a const reference, you cannot accidentally pass a temporary for the output of a function. I find this is a good thing, but some disagree. Quote: However, it is legal to call any member function on a temporary. And the line above is equivalent to: Foo(300).operator=( Foo(500) ); On the other hand, it is sometimes convenient (and somewhat less error prone) to write something like: getSomeTermporaryHandle().doSomething(); And there is no special treatment for assignment operators, which is just another member function. I think a balance had to be chosen between convenence and safety. Ivan -- http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form Back to top Matthew Schaefer Guest Posted: Sun Jan 30, 2005 11:16 am    Post subject: Re: Foo(300) = Foo(500); Why does a compiler compile that? I think the problem is the following: 1) Foo(300) is non-const. 2) your operator= function does nothing. It simply assigns the lhs back to itself. It does not check that the members of lhs and rhs are equal. 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