Summary: std.array.put doesn't put--it takes.
           Product: D
           Version: 2.030
          Platform: x86
        OS/Version: Windows
            Status: NEW
          Severity: major
          Priority: P2
         Component: Phobos

--- Comment #0 from Paul D. Anderson <> 2010-02-27 
16:52:12 PST ---
>From the description of the put primitive in std.range:

"r.put(e) puts e in the range (in a range-dependent manner) and advances to the
popFront position in the range. Successive calls to r.put add elements to the
range. put may throw to signal failure."

>From the example of std.array for the put function:

void main()
    int[] a = [ 1, 2, 3 ];
    int[] b = a;
    assert(a == [ 2, 3 ]);
    assert(b == [ 5, 2, 3 ]);

So, "putting" 5 into the array a removes the first element in a, and changes
the value of the first element of b. I would expect the first assert in the
code above to read:

    assert(a == [ 5, 1, 2, 3 ]);

The implementation of std.array.put doesn't make sense: 

void put(T, E)(ref T[] a, E e) { assert(a.length); a[0] = e; a = a[1 .. $]; }

It modifies a[0] and then replaces the array with the tail of the array,
omitting the first element.

It's possible there is some arcane meaning to the word "put" that I'm not aware
of, but if it means "insert an element at the front of the range" then
std.array.put is wrongly implemented.


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