Re: i+++j

[Ben Abbott]

On Feb 29, 2012, at 3:48 PM, Muhali wrote:

> In pushing my math limits, I stumbled upon
> 
> i+++j
> 
> octave apparently does this:
> 
> octave:> i=j=0
> i = 0
> octave:> k=i+++j; disp([i j k]) ;
>   1   0   0
> octave:> k=i+++j; disp([i j k]) ;
>   2   0   1
> 
> that is, doing first i++ and then adding j which always remains 0. It could
> be the other way round (leaving i and increasing j). So, is there a
> rationale for this, and should it not be as in matlab which does nothing,
> not even complain, no matter how many additional +'s are used?
> 

Octave is working as intended. The following is parsed the same way.

        i=j=0
        i = 0
        j
        j = 0
        k = (i++) + j; disp ([i j k]) ;
           1   0   0
        k = (i++) + j; disp ([i j k]) ;
           2   0   1

Ben