Without assigning it to a variable, I don't see what sense the
statement even makes.

 (⊃⊃x)

produces an intermediate value.  I can't think of what sense making an
assignment to an intermediate value even does.

I think APL should just throw a syntax error as in:

      (+/10 10⍴⍳100)[2]←4
SYNTAX ERROR
      (+/10 10⍴⍳100)[2]←4
      ^                ^

(Surely a syntax error is more appropriate than a segfault!)

What does IBM APL do?

--blake


On Sat, Apr 4, 2020 at 5:03 PM Rowan Cannaday <cannad...@gmail.com> wrote:

> hello y'all, hope everyone is staying safe.
>
>       x ← 1 (2 3 (4 5))
>       (⊃⊃x)[2;3;] ← 6 7
>
>
> ===================================================
> SEGMENTATION FAULT
>
> ----------------------------------------
> -- Stack trace at main.cc:88
> ----------------------------------------
> 0x7F032D7BABBB __libc_start_main
> 0x557F8B045425  main
> 0x557F8B1BE755   Workspace::immediate_execution(bool)
> 0x557F8B099A4B    Command::process_line()
> 0x557F8B09A45A     Command::do_APL_expression(UCS_string&)
> 0x557F8B099AE8      Command::finish_context()
> 0x557F8B0A3698       Executable::execute_body() const
> 0x557F8B159D4C        StateIndicator::run()
> 0x557F8B0DB23D         Prefix::reduce_statements()
> 0x557F8B0DA17D          Prefix::reduce_MISC_F_B_()
> 0x557F8B0F1F4F           Bif_F12_PICK::eval_B(Value_P)
> 0x557F8B0EBB33            Bif_F12_PICK::disclose(Value_P, bool)
> 0x557F8B0EB86A             Bif_F12_PICK::compute_item_shape(Value_P, bool)
> 0x7F032DCD1520
> 0x557F8B04C4AB
> ========================================
> ====================================================
>
> By the way, it works if you use an intermediary variable:
>
>       y←⊃⊃x
>       y[2;3;]
> 4 5
>       y[2;3;] ← 5 6
>       y
> 1 0
> 0 0
> 0 0
>
> 2 0
> 3 0
> 5 6
>
> Cheers,
>  - Rowan
>

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