[racket] Why choose the 'let*' construct over the 'define' construct when both can have sequential scope?

Because of this: http://docs.racket-lang.org/guide/set_.html#%28part._using-set%21%29
Would this be better:
(thread-through x e …)
==
(let* ([x e] …) x)

On Feb 19, 2015, at 1:37 PM, Matthias Felleisen <matthias at ccs.neu.edu> wrote:

> 
> I recommend a macro for these situations, it's almost always what you want: 
> 
> (thread-through x e_0 e ...)
> ==
> (let ([x e_0])
>  (set! x e)
>  ...
>  x)
> 
> (you can also use let* here). 
> 
> This brings across what you're really doing. 
> 
> 
> 
> On Feb 19, 2015, at 1:32 PM, Jens Axel Søgaard <jensaxel at soegaard.net> wrote:
> 
>> We have let and let*, but only define. I miss define* at times.
>> 
>> This is an error:
>> 
>>   (block
>>     (define x 4)
>>     (define x (+ x 1))
>>     (+ x 2))
>> 
>> With a define* it becomes:
>> 
>>   (block
>>     (define x 4)
>>     (define* x (+ x 1))
>>     (+ x 2))
>> 
>> which (should) expand to:
>> 
>>   (block
>>     (define x 4)
>>     (block
>>       (define x (+ x 1))
>>       (+ x 2)))
>> 
>> /Jens Axel
>> 
>> 2015-02-19 19:24 GMT+01:00 Laurent <laurent.orseau at gmail.com>:
>>> But in such situations you can be vicious and abuse `let` to have your
>>> `define`s:
>>> (let ()
>>> (define x 3)
>>> (define y 4)
>>> (list x y))
>>> 
>>> It's even more vicious if you use `let*` instead of `let`, but quite less if
>>> you use `begin` instead ;)
>>> 
>>> 
>>> 
>>> On Thu, Feb 19, 2015 at 5:49 PM, Matthias Felleisen <matthias at ccs.neu.edu>
>>> wrote:
>>>> 
>>>> 
>>>> In some places, you are allowed only one expression, and for that
>>>> situation, you need let*.
>>>> 
>>>> 
>>>> On Feb 19, 2015, at 12:40 PM, Don Green <infodeveloperdon at gmail.com>
>>>> wrote:
>>>> 
>>>>> What is/are the reason(s) for choosing the 'let*' construct over the
>>>>> 'define' construct?
>>>>> 
>>>>> (define (print-two f)
>>>>> (let* ([_ (print (first f))]
>>>>>        [f (rest f)]
>>>>>        [_ (print (first f))]
>>>>>        [f (rest f)])
>>>>>   f))
>>>>> 
>>>>> (define print-two
>>>>> (lambda (f)
>>>>>  (print (first f))
>>>>>  (set! f (rest f))
>>>>>  (print (first f))
>>>>>  (set! f (rest f))
>>>>>  f))
>>>>> 
>>>>> (void (print-two '(1 2))) ;=> 12
>>>>> 
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>>> 
>>> 
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>> 
>> 
>> 
>> -- 
>> --
>> Jens Axel Søgaard
> 
> 
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