Re: Example where Euphoria defaults on a boolean expression

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Rom wrote:

>    if (f = e) and ( f != e) then print( 1, 1)
>    else print ( 1, 0)
>    end if
>
>
>f = e and f != e cannot be resolved.  The evaluation should return nil.
>
>  
( f = e ) and ( f != e) evaluates to 0 also.
( 3 = 2) and (3 != 2) => 0
0 and 1 => 0

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