Saaa wrote:
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.
You cannot have static array be ref or out. You must use int[] arr
instead and let the caller specify the length.
Array in D are already references to their data, it's a 8bytes (or
16bytes on x64) value containing a pointer and a length. So the
following prototype:
void func(<none>/in/ref/out int[] arr);
would have the following semantics:
"<none>" copies the array reference, the referenced data is mutable.
Modifying the local reference does not change the caller's reference.
"in" copies the array reference, the local reference AND the referenced
data are immutable in the method's scope.
"ref" passes a reference to the caller's array reference, the referenced
data is mutable. Modifying the local reference also changes the caller's
reference.
"out" passes a reference to the caller's array reference. The referenced
array reference is zeroed and can be modified by the local reference.
If you want a mutable reference to an immutable view on the referenced
data, use const(int)[], which can also be ref or out.
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
"<none>" would be a mutable copy.
"in" would be immutable copy.
"out" is a reference to the caller's int initialized to 0.
"ref" is a reference to the caller's int.
"in" and "const" are only really useful with types which are already
references, such as pointers, arrays and objects.
Jeremie
