Re: for loop 'bad math expression'

[Ray Andrews <rayandrews@xxxxxxxxxxx>]

On 2024-02-04 06:43, Mark J. Reed wrote:

That's not qutie what's happening. Arithmetic expressions interpret all variables as numbers, whether the variables are declared as integers or not. If a variable's value doesn't look like a number, it is interpreted as the name of another variable, which is then looked up recursively. If it gets to a variable that doesn't exist, the value is taken as 0.

    I get that.  $(( abc ..... )) ... abc isn't a number therefore it's
    the name of a variable.  However I did:
    
    % var=abc
    ... outside any (()) and the conversion happened 'retroactively' so
    to speak. 
    
    % var=path; print -- $var
    path
    ... 'path' is just a string, there's no implicit conversion to a
    variable.  Ergo:
    
    %var=abc 
    ... seems to me the same.  Later on abc becomes an integer name, and
    the string is lost

... it will be converted back to a string that just happens to be all digits. What's the difference? Well, let's continue your example:

zsh% var=abc; let var+=2; echo $var

2

zsh% var+=2; echo $var # note: no let

22

    Right, I get that too.  A string of digits is still a string.  
    
    Anyway it goes back to my original
      speculation/question that implicit promotions/conversions do
      happen thus a 'demotion' -- in the case of that glob expansion --
      is something that might at least be contemplated even if in fact
      the idea is unsound.  As for me I think that taking a string and
      then using it as the name of an integer is a bit strange, but it
      is what it is and I won't crash into that problem again.