On Thursday, 25 June 2020 at 03:35:00 UTC, repr-man wrote:
I have the code:
int[5] a = [0, 1, 2, 3, 4];
int[5] b = [5, 6, 7, 8, 9];
auto x = chain(a[], b[]).chunks(5);
writeln(x);
It produces a range of slices as is expected: [[0, 1, 2, 3, 4],
[5, 6, 7, 8, 9]]
However, when I define a function as follows and pass in the
result of the chain iterator:
auto func(R)(R r, size_t width)
if(isRandomAccessRange!R)
{
return r.chunks(width);
}
void main()
{
int[5] a = [0, 1, 2, 3, 4];
int[5] b = [5, 6, 7, 8, 9];
auto x = func!(int[])(chain(a[], b[]), 5);
writeln(x);
}
It gives me an error along the lines of:
Error: func!(int[]).func(int[] r, ulong width) is not callable
using argument types (Result, int)
cannot pass argument chain(a[], b[]) of type Result to
parameter int[] r
I was hoping it would return the same result as the first
program.
This seems to have to do with the fact that all iterators
return their own unique type. Could someone help me understand
the reason behind this design and how to remedy my situation?
Chain returns a range not an int[]. You need to either convert
the range to an array via .array or allow the compiler to infer
the type of the parameter of func (You'll need to import
std.range to have the range interface available)
mhh
