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Re: ReplaceAll doesn't replace


In a message dated 1/26/02 4:37:21 AM, kmorga51 at calvin.edu writes:

>I understand how the following will replace x with the list
>
>#^2 & /@ (x /. x -> {a, b, c})
>
>to generate
>
>{a^2, b^2, c^2}
>
>But, why isn't x replaced at the beginning of the evaluation in the
>following
>
>(#^2 & /@ x) /. x -> {a, b, c}
>
>since it generates
>
>{a, b, c}
>
>What I really want to know is: What is it about the Function function that
>doesn't allow ReplaceAll to "replace all" at the beginning of an evaluation?
>

Because

(#^2 & /@ x)

x

There was no list so there was no mapping.  To have the mapping occur, you 
would need to use

(#^2 & /@ {x}) /. x -> {a, b, c}

{{a^2, b^2, c^2}}


Bob Hanlon
Chantilly, VA  USA


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