Re: for loop 'bad math expression'

["Mark J. Reed" <markjreed@gmail.com>]

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On Sat, Feb 3, 2024 at 8:14 PM Ray Andrews <rayandrews@eastlink.ca> wrote:

> 0 /usr/share/info 0 % var=abc; let var+=2; echo $var
> 2
>
> ... it looks like the shell is simply throwing away 'abc' and starting
> afresh with an integer and then incrementing it.
>

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.

So, when you do any sort of assignment inside a *let* or *((*...*))* or
array subscript or any other arithmetic context, the variable being
assigned will *always* come out of it with a numeric value. But unless
you've done a *declare -i *or *typeset -i*, it will not be stored as an
integer; 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*


As you can see, outside of *let*, using *+=* *appended* to the variable
instead of doing arithmetic. Because it's still a string.

If you were to go in and declare it as an integer, then *+= *would have its
arithmetic meaning even outside of explicit arithmetic context:

*zsh% *typeset -i var

*zsh% *var+=2; echo $var
*24*


But that's only because of the *typeset*. It doesn't happen automatically.

-- 
Mark J. Reed <markjreed@gmail.com>

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