Does specifying the number of elements on defining a vector in Racket make the program use less time-CodePudding

For example, can this

#!/usr/bin/env racket
#lang racket/base 

(define vector-of-pleasures #6("eating" "cold showers" "swimming" "running" "pullups" "weighlifting"))
(for ((pleasure vector-of-pleasures)) (displayln pleasure))

be more performant because of the optional annotation than this

#!/usr/bin/env racket
#lang racket/base

(define vector-of-pleasures #("eating" "cold showers" "swimming" "running" "pullups" "weighlifting"))
(for ((pleasure vector-of-pleasures)) (displayln pleasure))

Methinks it should not matter since in both snippets vector-of-pleasures is immutable. What are, other than performance, possible reasons for annotating the number of the elements of the vector on its definition?

CodePudding user response：

No, it doesn't. The real use of that syntax is that it lets you write a vector of a bunch of identical things easily: #6(1) is a vector of 6 1s.

But I think it could save space. I can't see why, for instance:

(let ((v #2((1 2))))
  (eq? (vector-ref v 0) (vector-ref v 1)))

should not be true. But then this rapidly turns into the question of when similar literals can be folded: can (eq? '(1 2) '(1 2)) return true for instance? I don't know what Racket's rules are on that.