Why D const is annoying

Steven Schveighoffer schveiguy at yahoo.com
Mon Dec 12 04:50:35 PST 2011

On Sun, 11 Dec 2011 12:07:37 -0500, Mafi <mafi at example.org> wrote:

> Am 10.12.2011 21:25, schrieb Walter Bright:
>> On 12/10/2011 11:03 AM, Mehrdad wrote:
>>> So how are you supposed to implement opApply on a container (or e.g.
>>> here, a
>>> matrix)? Copy/paste the code for const- and non-const versions?
>> Internal to a function, inout behaves like 'const'. You won't be able to
>> modify the data. Therefore, if there is no inout in the return type, use
>> 'const' in the parameter list instead.
>> The purpose of inout is to transmit the 'constness' of the function
>> argument type to the return type, using only one implementation of that
>> function. That requires the function to internally regard inout as  
>> const.
> But what about:
> void f(ref inout(int)* a, inout(int)* b) { a = b; }
> This cant work with const because that would violate the const system. I  
> think the rule should be that either the return type must be inout or at  
> least one ref/out parameter.
> Am I overlooking something?

That was brought up during discussion on adding the feature.  One of the  
reasons inout is viable is because a) the source and result of where the  
constancy flows is well defined and b) the exit point is an rvalue

Allowing ref parameters fails both those rules.

Consider this:

void bad(ref inout(int)* a, ref inout(int)* b);

which is the entry and which is the exit?  Is a set to b, or b set to a?

Now, also consider that you can't affect the constancy of the result,  
because the type of the parameter is already defined.  e.g.:

// note that your example shouldn't even be valid, because you can't  
implicitly cast through two mutable references
int a;
auto pa = &a;
immutable int b;
auto pb = &b;

f(a, b);

How can this affect a?  its type is already decided.  Compare that to:

inout(int)* g(inout(int)* b) { return b;}

auto pa = g(pb);

clean and simple.


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