Why does enumerate over range return dchar, when ranging without returns char?
rikki cattermole
rikki at cattermole.co.nz
Thu May 3 05:56:26 UTC 2018
On 03/05/2018 5:44 PM, James Blachly wrote:
> I am puzzled why enumerating in a foreach returns a dchar (which forces
> me to cast), whereas without the enumerate the range returns a char as
> expected.
>
> Example:
>
> ```
> import std.stdio;
> import std.range : enumerate;
>
> void main()
> {
> char[] s = ['a','b','c'];
>
> char[3] x;
> auto i = 0;
> foreach(c; s) {
> x[i] = c;
> i++;
> }
>
> writeln(x);
> }
> ```
> Above works without cast.
>
> '''
> import std.stdio;
> import std.range : enumerate;
>
> void main()
> {
> char[] s = ['a','b','c'];
>
> char[3] x;
> foreach(i, c; enumerate(s)) {
> x[i] = c;
> i++;
> }
>
> writeln(x);
> }
> ```
> Above fails without casting c to type char.
>
> The function signature for enumerate shows "auto" return type, so that
> does not help me understand.
>
> Kind regards
The first example uses auto-decoding (UTF-8 codepoints into a single
UTF-32 one). This is considered a bad thing. But the compiler can
disable it and leave it as UTF-8 code point upon request.
The second example returns a Voldemort type (means no-name) which
happens to be an input range. Where it can't disable anything and has
been told that it is returning a dchar. See[0] as to where this gets
decoded. Writing two small functions to replace it (and popFront), will
override this behavior.
[0] https://dlang.org/phobos/std_range_primitives.html#.front
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