error with std.range.put - readN

[Jonathan M Davis via Digitalmars-d]

On Monday, 4 May 2015 at 14:33:23 UTC, Baz wrote:
> the following program fails because of the `put` function :
>
> ---
> import std.stdio;
> import std.range;
>
> size_t readN(T, Range)(ref Range src, ref Range dst, size_t n)
> if (isInputRange!Range && isOutputRange!(Range, T))
> {
>     size_t result;
>
>     while(1)
>     {
>         if (src.empty || result == n)
>             break;
>
>         put(dst, src.front()); // here
>         src.popFront;
>
>         ++result;
>     }
>
>
>     return result;
> }
>
> void main(string[] args)
> {
>     int[] src = [1,2,3];
>     int[] dst = [0];
>
>     auto n = readN!int(src, dst, 2);
>
>     writeln(dst);
> }
> ---
>
> If i replace `put` by a cat op (`~`) it works, however the cat 
> op only works here because i test the template with two int[].
>
> What's wrong ?

Your destination is too small. When arrays are treated as output 
ranges, they get written to like a buffer. They don't get 
appended to. If you want to append to them, then use Appender for 
the output range.

On a side note, it's very backwards that you're calling front 
with parens and popFront without. front is usually a property 
function (in which case, if @property ever gets sorted out 
properly, front() wouldn't compile), and popFront is never a 
property function (so you _can_ call it without parens, but it's 
kind of weird to do so).

- Jonathan M Davis