error with std.range.put - readN

[sclytrack 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 ?

I believe the put(R,E) calls the doPut(R, E)


private void doPut(R, E)(ref R r, auto ref E e)
{
//...
  else static if (isInputRange!R)
{
    static assert(is(typeof(r.front = e)),
         "Cannot nativaly put a " ~ E.stringof ~ " into a " ~ 
R.stringof ~ ".");
    r.front = e;
    r.popFront();
}
//...
}


So basically you put elements into the list until it is empty.


import std.stdio;
import std.range;

void main()
{
    int [] list = new int [10];
    writeln(list);    //10 zeros
    list.front = 5;   //1 five and 9 zeroes
    writeln(list);
    list.popFront();  //9 zeroes left.
    writeln(list);
}