Invariant Question (yes, another one)

Reiner Pope some at address.com
Thu Nov 15 15:41:13 PST 2007


Janice Caron wrote:
> On 11/15/07, Steven Schveighoffer <schveiguy at yahoo.com> wrote:
>> Your first data type is an AA with invariant strings as keys and values.
>> HOWEVER, the AA itself is not invariant.  Suppose after you converted to an
>> AA with const strings as keys and values, you replaced one of the values
>> with a mutable string cast into a const string.
> 
> Yes, you're right.
> 
> invariant(char)[][] won't implicitly convert to const(char)[][]
> either, for the same reason.
> But...
> invariant(char[])[] will implicitly conver to const(char[])[]
> 
> But here's the odd thing...
> 
> invariant(char[])[] will implicitly conver to const(char[])[]
> but
> invariant(char[])[int] won't implicitly convert to const(char[])[int]
> 
> The only difference is the int in the square brackets.

I think it's just the odd behaviour that invariant(char[])[] is actually 
the same as invariant(char[][]), which is implicitly convertible to 
const(char[][]), the same as const(char[])[].

   -- Reiner



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