On 12/5/12 6:07 PM, Niko Matsakis wrote:
Great article. Regarding why ~[T] and ~T behave somewhat differently, the reason is that the size of [T] is not known to the compiler! It's kind of like how in C, if you have T[3], that's a perfectly fine type, but T[] is sort of a degenerate type that gets converted to T* willy-nilly. The reason is that the compiler can't really manipulate an "unknown number of T's in a row", which is what [T] (Rust) and T[] (C) represent. As an example, it can't put such a value on the stack, since it doesn't know how much stack space to allocate (well, it can't put that on the stack without using alloca() or something similar).
We have talked about special-casing "&*" in the grammar to mean "borrow" regardless of the type of pointer, and we could also conceivably special-case "@*" and "~*", which would make what you wrote work. Basically you would be permitted to dereference &[T]/@[T]/~[T]/&str/@str/~str for the sole purpose of borrowing it or copying its contents into another heap.
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