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 | Posted by Jonathan M Davis |  Permalink Reply |
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Jonathan M Davis 
| http://d.puremagic.com/issues/show_bug.cgi?id=6289
Summary: Make slices of const/immutable arrays mutable (but
keep the elements const/immutable)
Product: D
Version: unspecified
Platform: All
OS/Version: All
Status: NEW
Severity: enhancement
Priority: P2
Component: DMD
AssignedTo: nobody@puremagic.com
ReportedBy: jmdavisProg@gmx.com
--- Comment #0 from Jonathan M Davis <jmdavisProg@gmx.com> 2011-07-11 17:12:35 PDT ---
Okay. If you have
immutable val = [1, 2, 3, 4];
void func(immutable(int)[] arg) {}
you can pass val to func even though val is fully immutable. The compiler realizes that it's safe, because func cannot alter the original array. The array in the function is a slice of the original and can't affect the original (since the elements are immutable). However, even though this is true, the type of a slice of an immutable int[], is still immutable int[], not immutable(int)[] as with the function parameter. I'd like to see
immutable val = [1, 2, 3, 4];
assert(is(typeof(val) == immutable(int[])));
assert(is(typeof(val[]) == immutable(int)[]));
The slice is then mutable (though its elements are not) - which makes sense, since that's exactly what the function described above does. The reason that I want this to be the case is that you can then use immutable arrays with range-based functions if you slice them (just as is the case with static arrays). As it stands, both
find(val, 3);
and
find(val[], 3);
are illegal, because the compiler tries to instantiate find with immutable(int[]). But if the type of a slice of val were immutable(int)[], then the second find call would work. This would help solve the problem of immutable arrays not working with range-based functions (see bug# 6148). Then, if you slice static arrays and immutable/const arrays (just as you'd have to do with container), they will work with range-based functions. As it stands, that works with static arrays but not immutable or const arrays.
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