ref const array error

[jdrewsen]

On Wednesday, 18 January 2012 at 23:09:56 UTC, Timon Gehr wrote:
> On 01/18/2012 10:12 PM, jdrewsen wrote:
>> On Wednesday, 18 January 2012 at 20:13:04 UTC, Timon Gehr 
>> wrote:
>>> On 01/18/2012 08:59 PM, jdrewsen wrote:
>>>> On Wednesday, 18 January 2012 at 19:43:52 UTC, Timon Gehr 
>>>> wrote:
>>>>> On 01/18/2012 08:31 PM, jdrewsen wrote:
>>>>>> Recently the encoding.safeDecode stopped working for some 
>>>>>> of my
>>>>>> existing
>>>>>> code. This example outlines the issue:
>>>>>>
>>>>>> import std.encoding;
>>>>>>
>>>>>> void main(string[] args) {
>>>>>> auto e = EncodingScheme.create("utf-8");
>>>>>> auto a = new byte[100];
>>>>>> e.safeDecode(a);
>>>>>> }
>>>>>>
>>>>>> Results in:
>>>>>>
>>>>>> Error: function std.encoding.EncodingScheme.safeDecode (ref
>>>>>> const(ubyte)[] s) const is not callable using argument 
>>>>>> types (byte[])
>>>>>>
>>>>>> Isn't this an error in the compiler?
>>>>>>
>>>>>> /Jonas
>>>>>>
>>>>>
>>>>> No, this is a bugfix. The operation is unsound:
>>>>>
>>>>> immutable(ubyte)[] foo(ref const(ubyte)[] s){
>>>>> auto r = new immutable(ubyte)[1];
>>>>> s = r;
>>>>> return r;
>>>>> }
>>>>>
>>>>> void main() {
>>>>> ubyte[] x;
>>>>> immutable(ubyte)[] y = foo(x);
>>>>> static assert(is(typeof(y[0])==immutable));
>>>>> auto oldy0 = y[0];
>>>>> x[0]=oldy0+1;
>>>>> assert(oldy0 == y[0]); // fail
>>>>> }
>>>>>
>>>>> The functionality is not going away; You will be able to 
>>>>> use inout for
>>>>> the same purpose once my enhancement request gets 
>>>>> implemented:
>>>>> http://d.puremagic.com/issues/show_bug.cgi?id=7105
>>>>
>>>> Wouldn't a nicer solution be to let the compiler ensure that
>>>> an immutable array cannot escape through a ref const array 
>>>> parameter?
>>>>
>>>> /Jonas
>>>>
>>>
>>> That would not suffice.
>>>
>>> ubyte[] foo(ref const(ubyte)[] s){
>>> auto r = new ubyte[1];
>>> s = r;
>>> return r;
>>> }
>>>
>>> void main() {
>>> immutable(ubyte)[] x;
>>> ubyte[] y = foo(x);
>>> static assert(is(typeof(x[0])==immutable));
>>> auto oldx0 = x[0];
>>> y[0]=oldx0+1;
>>> assert(oldx0 == x[0]); // fail
>>> }
>>
>> In the example foo is actually using the ref s parameter as an 
>> out
>> parameter. The compiler could catch that you're doing this and 
>> show an
>> error.
>>
>> This would force you to let foo look like:
>>
>> ubyte[] foo(out const(ubyte)[] s);
>>
>> Wouldn't that fix it?
>>
>>
>
> If it is ref or out is irrelevant for the example, so how would 
> this fix anything? The compiler could, in principle, treat 
> const similarly to inout (just without the context sensitivity 
> and parameter matching etc) for 'ref' parameters and do all the 
> type checking at the call site. However, that would then 
> restrict what the callee can do and introduce a strange special 
> case. inout is the way to go.

But in the example you're using s as an out parameter and that 
should trigger the error I got originally of course. But if ref 
parameters were disallowed to be used as out parameters the 
compiler would catch the error in your example wouldn't it?