On 4/25/22 16:59, Salih Dincer wrote:
> Because it cannot be used with other possibilities such as
> ```chunks()``` and ```take()```.
There must be something wrong. map is commonly used with chunks(),
take(), etc.
> Also it cannot be customized with
> ```toString()```.
Can you show examples of these please?
> I guess even when ```this()``` constructor is added,
> ```map()``` explodes!
Ah! Now I tried *removing* Bar.this and the compilation failed. Perhaps
you left the working code in? (?)
For others:
1) Remove Bar.this
2) The code will fail below:
auto arr2 = "abcdefghi"
.chunks(3)
.map!(a => Bar(a)) <-- ERROR
.array;
Error: cannot implicitly convert expression `a` of type `Take!string` to
`string`
One confusing thing is the fact that we learn that chunks uses take in
its implementation. I think it's cool but a newcomer may be confused
with where that Take comes from. Let's put that aside.
Then the compilation error is easy to understdand because chunks returns
a range itself but it is not possible to make a Bar from a range. I
don't agree that this is a problem with map's usability. The type system
doesn't know what to do.
Here is one way of fixing the issue, not surprisingly, with map itself. ;)
import std.conv : to;
auto arr2 = "abcdefghi"
.chunks(3)
.map!(c => c.to!string) // <-- ADDED
.map!(a => Bar(a))
.array;
Or, one can use a single map expression:
auto arr2 = "abcdefghi"
.chunks(3)
.map!(a => Bar(a.to!string)) // <-- COMBINED
.array;
Or, a function can be called, etc.
But to remove a misunderstanding, map can be used with most other range
algorithms, in the standard library, provided by the programmer, etc.
Ali
