Issue 6922 - [TDPL] superimposing of const and immutable does not work correctly #505
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| static assert(is( inout(immutable(T)) == immutable(T) )); | ||
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| static assert(is( inout(const(T)) == const(T) )); | ||
| static assert(is( const(inout(T)) == const(T) )); |
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These two lines are not correct. const cannot override inout, because inout could mean immutable and immutable overrides const. inout cannot override const because inout could mean unqualified. Therefore inout(const(T)) and const(inout(T)) are effectively irreducible. (But the compiler should prefer one of them to the other, probably inout(const(T)).)
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Current dmd implementation does not allow that T has two modifiers const and inout at the same time.
And inout(T) is implicitly convertible to const(T), then I have thought inout(const(T)) should parsed as const(T).
Yes, this is an arbitrary rule. Instead, we can also select inout(T) as the parsing result of inout(const(T)).
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Oh, I see. That obviously also complicates the implementation of
http://d.puremagic.com/issues/show_bug.cgi?id=6930
Because there would have to be AST nodes for inout(const(T)) and shared(inout(const(T)).
See discussions of bug 6922 and 6930.
workaround Bug 2962
http://d.puremagic.com/issues/show_bug.cgi?id=6922