Not more efficient, but perhaps more readable:
(define (repeater f count)
(for ([x (in-range count)])
(f)))
On Mon, Sep 15, 2014 at 12:50 PM, George Rudolph <rudolp...@citadel.edu>
wrote:
> All,
>
> I am sure this question has been answered before, but I can’t find it.
>
> Suppose I want to call some function, either built-in or user-defined, n
> number of times, and measure the total time it takes. Is there a more
> efficient or elegant way
>
> of doing this than passing the expression to a tail-recursive counting
> loop such as the following?
>
>
>
> (define (repeater sexpr count)
>
> (sexpr)
>
> (cond
>
> [(> count 0) (repeater sexpr (- count 1)) ])
>
> )
>
>
>
>
>
>
>
> George Rudolph
>
> Associate Professor of Computer Science
>
> Department of Mathematics and Computer Science
>
> 225 Thompson Hall
>
> The Citadel
>
> 171 Moultrie Street
>
> Charleston, SC 29409
>
> 843.953.5032
>
>
>
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