On 2010-08-06 19:33, Philippe Sigaud wrote:
2010/8/6 Adrian Matoga <[email protected] <mailto:[email protected]>>
Hi,
Is there any off the shelf solution for iterating over a range by
chunks?
None that I know of.
(should substitute [1, 2, 3], [4, 5, 6], [7, 8, 9], [10] for chunk
in subsequent iterations)
As a data point, why do you think it should produce [10] and not stop at
[7,8,9]?
By an analogy to taking a file by chunks. The other case also occured to
me but I simply thought I could imagine more situations in which I need
the whole input range to be exhausted, including remainder.
Here is what I cooked, it's still a bit rough around the edges. It has
an optional step argument, to see how many elements to jump.
import std.range;
struct Chunks(R) if (isInputRange!R)
{
R range;
size_t n; // chunk size
size_t step; // how many elements to jump
bool empty() @property { return range.empty;}
ElementType!R[] front() @property { return array(take(range, n));}
// inefficient if you call front() many times in a row
void popFront() @property { popFrontN(range, step);}
static if (hasLength!R)
size_t length() @property { return (range.length+step-1)/step;}
}
Chunks!R chunks(R)(R range, size_t n) if (isInputRange!R)
{
return Chunks!R(range, n, n); // default is step == n
}
Chunks!R chunks(R)(R range, size_t n, size_t step) if (isInputRange!R)
{
return Chunks!R(range, n, step);
}
Thank you very much!
And yes, I also think it should be included in std.range (actually
before posting the question I was almost sure I could find it there ;)).
Adrian
