Steven Schveighoffer wrote: > On Mon, 21 Sep 2009 11:12:49 -0400, #ponce <alil...@gmail.com> wrote: > >> Is there a reason to use ref instead of in, inout or out ? >> I'm using D1 and it's not in the spec.
It is here in the spec: Function Parameters http://www.digitalmars.com/d/1.0/function.html > > ref and inout are synonymous, ref is preferred, as inout is essentially a > defunct keyword. > > in means that it's a reference that cannot be changed. In D1, it means > you cannot change the value, but if the value contains a reference to > something else (like an array), you can change what it points to. The spec says: (http://www.digitalmars.com/d/1.0/function.html) "For dynamic array and object parameters, which are passed by reference, in/out/ref apply only to the reference and not the contents." And: (http://www.digitalmars.com/d/1.0/abi.html) "When passing a static array to a function, the result, although declared as a static array, will actually be a reference to a static array." Shouldn't it thus be: "For array and object parameters, which are passed by reference..." ?? My interpretation for: void func(in/out/ref int[10]arr) "in" arr is a new reference pointing to the passed array. "out" arr is a new array of 10 ints initialized to 0. "ref" arr is the original reference you passed Same for dynamic arrays/objects. > In D2, it is synonymous with ref const scope (although the scope > attribute doesn't yet mean anything), so you cannot change anything it > points to. > > out means that it's a reference that you are always going to overwrite. I > believe the compiler even initializes the value for you upon function > entry. > > So in summary: > > inout: don't use it. > ref: Use this for pass by reference for a value that you may or may not > change. > in: Use this for pass by reference for a value that you won't change > out: Use this for pass by reference for a value that you will *always* > change. What do you mean by: "pass by reference for a value"? It reads like it creates a reference to a value. in and out don't create a reference, do they? My interpretation for: void func(in/out/ref int i) "in" i is a new int initialized to the passed int. "out" i is a new int set to 0. "ref" i is a reference to the original passed int? Are my interpretations correct?? Anyways, I think this deserves a bit more explaination in the spec. Ref is never explained. > > -Steve
