June 16, 2016 const types can't be specialized non-const, if arrays? | ||||
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I don't get it. Do I have to write a separate template for arrays specifically or something? NonConst foo(Constant: const NonConst, NonConst)(Constant bar) { pragma(msg,"NonConst is ",NonConst); pragma(msg,"Constant is ",Constant); NonConst foo = bar; return foo; } void main() { const int bar = 42; auto baz = foo(bar); pragma(msg,typeof(baz)); baz = 23; const(int[]) barr = [1,2,3]; auto bazz = foo(barr); pragma(msg,typeof(bazz)); bazz[0] = 4; } /* NonConst is int Constant is const(int) int NonConst is const(int)[] Constant is const(int)[] const(int)[] derp.d(16): Error: cannot modify const expression bazz[0] Failed: ["dmd", "-v", "-o-", "derp.d", "-I."] */ |
June 16, 2016 Re: const types can't be specialized non-const, if arrays? | ||||
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Posted in reply to cy | On 6/16/16 3:43 PM, cy wrote:
> I don't get it. Do I have to write a separate template for arrays
> specifically or something?
>
> NonConst foo(Constant: const NonConst, NonConst)(Constant bar) {
> pragma(msg,"NonConst is ",NonConst);
> pragma(msg,"Constant is ",Constant);
> NonConst foo = bar;
> return foo;
> }
>
> void main() {
> const int bar = 42;
> auto baz = foo(bar);
> pragma(msg,typeof(baz));
> baz = 23;
> const(int[]) barr = [1,2,3];
> auto bazz = foo(barr);
> pragma(msg,typeof(bazz));
> bazz[0] = 4;
>
> }
>
> /*
>
> NonConst is int
> Constant is const(int)
> int
> NonConst is const(int)[]
> Constant is const(int)[]
> const(int)[]
> derp.d(16): Error: cannot modify const expression bazz[0]
> Failed: ["dmd", "-v", "-o-", "derp.d", "-I."]
>
> */
>
1. Because const(int)[] is not the same as const(int[]). The former is a *mutable* array of *constant* integers.
2. Yes, I see that you passed in const(int[]). IFTI was changed a while back to automatically convert to tail-const versions when implying parameters (to save on template bloat). So it deduces "Constant" to be const(int)[].
Also, note that you cannot assign a const item containing references to a non-const version of it. So even if NonConst was properly deduced, the function would fail to compile.
-Steve
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