Why is pop faster than shift?

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Douglas Crockford, in JavaScript: The Good Parts, states that "shift is usually much slower than pop". jsPerf confirms this. Does anyone know why this is the case? From an unsophisticated point of view, they seem to be doing pretty much the same thing.

javascript performance
6 Answers

The difference can be negligible—Unoptimized executors may run shift much slower than pop, but optimized ones won't.

You can optimize like this:

let WrapArray = _=>{
  //Ensure no other ref to `_`.

  let numlike = _=>isNaN(_)?false:true
  let num = _=>Number(_)
  {
    let shift_q = 0
    return new Proxy(_, {
      get(first_t, k){
        switch(k){
          case 'shift': return (z={})=>(z.r=first_t[0 + shift_q], delete first_t[0 + shift_q++], z.r)
          break; case 'length': return first_t.length - shift_q
          break; default: return first_t[numlike(k)?num(k) +/*todo overflowguard*/shift_q:k]
        }
      },
      set(first_t, k, v){
        switch(k){
          case 'length': first_t.length = v + shift_q
          break; default: first_t[numlike(k)?num(k) +/*todo overflowguard*/shift_q:k] = v
        }
      }, 
      has(first_t, k){
        return (numlike(k)?num(k) +/*todo overflowguard*/shift_q:k) in first_t
      },
      deleteProperty(first_t, k){
        delete first_t[numlike(k)?num(k) +/*todo overflowguard*/shift_q:k];return 543
      },
      apply(first_t, t, s){
        first_t.call(t, s)
      },
      construct(first_t, s, t){
        new first_t(...s)
      },
    })
  }
}
(_=WrapArray(['a','b','c'])).shift()
console.log(_.length/*2*/, _[0]/*b*/, _[1]/*c*/, _[2]/*undefined*/)
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