http://gcc.gnu.org/bugzilla/show_bug.cgi?id=48483
--- Comment #13 from Jonathan Wakely <redi at gcc dot gnu.org> 2011-04-07 13:38:11 UTC --- (In reply to comment #10) > (In reply to comment #9) > > > No, the variable is in scope after its identifier, so it can be used in the > > initializer expression, e.g. > > > > int i = sizeof(i); // ok > > int i = i+1; // not ok > > My opinion is that C habit rules here. > > In C > > int i=i+1; -> int i; i=i+1; In C++ this can be written: int i(i+1); The initialization rules for C++ do > without constructors, objects e.t.c. > This example will bring just a warning. > > In C++ > > Class i(i.Method()); eval i.Method() -> construct with Class::Class(result) Note that this is similar to: Class i = i.Method(); For the purposes of your complaint it's not relevant that it's a class with a constructor. Obviously "i.Method()" has to be evaluated first at runtime, but that doesn't make it a compile-time error. "i" is in scope as soon a "Class i" has been parsed, so the expression "i.Method()" in the initializer can be compiled. > another order of calculation. (see comment 2) Must be an error. No. The behaviour at runtime is undefined, but it is not an error at compile time - the syntax is valid, but the result is undefined. I understand what you're saying, but you're wrong. C++ doesn't work how you think it works.
