Implementing pseudocode in JavaScript

Question

I'm quite new to JavaScript and I'm trying to implement my code in pseudocode in JavaScript, however I'm not getting the result that I'm supposed to. I want the function to permute the elements of the array p places to the left. In pseudocode I'm using a queue data structure, but I thought I can as well us an array. As the result of my function, I get an array with [2, 2, 2, 2]. Can you please help me out?

My code in pseudocode:

Function PERMUTEVECTOR(row, p)

    If p=0 then

        Return row

    End if

    New Queue q

    For 0<= i <4 do

        ENQUEUE[row[i], q]

    End for

 

    For 1<= i <= p do

        ENQUEUE[HEAD[q],q]

        DEQUEUE[q]

    End for

 

    For 0<=i<4 do

        Row[i] <- HEAD[q]

        DEQUEUE[q]

    End for

    Return row

End function

My code in JavaScript:

function permute_vector(row, p)

{

  if (p=0)

  {return row}

   

  let q = new Array()

  for (i=0; i<4; i++)

  {

    q.push(row[i])

  }

 

for (i=0; i<p; i++)

  {

    q.push(q[0])

    q.pop()

  }

 

  for (i=0; i<4; i++)

  {

    row[i] = q[0]

    q.pop()

  }

return row

}

 

px = permute_vector([2,4,1,3], 1)

console.log("px is:", px)

}

I did the same in Python and it works fine:

def permute_vector(row, p):

  if p==0:

    return row

  q = []

  for i in range(4):

    q.append(row[i])

  

  for i in range(p):

    q.append(q[0])

    q.pop(0)

  for i in range(4):

    row[i] = q[0]

    q.pop(0) 

  return row

What am I doing wrong with my JavaScript code?

Many thanks!

What's the purpose of all these line-feeds, besides having to scroll? — Martin Zeitler, Jun 26 '22 at 08:20
please add the wanted result. `=` is an assignment and in `if` ususally wrong ... — Nina Scholz, Jun 26 '22 at 08:24
programmers are not paid by the line - please make code readable — Bravo, Jun 26 '22 at 08:26
`As the result of my function, I get an array with [2, 2, 2, 2]` is that correct? — Bravo, Jun 26 '22 at 08:31
all that just to do `const q = row.slice(); while(p-- > 0) q.push(q.shift()); return q;`? or `row.push(...row.splice(0, p)); return row;` — Bravo, Jun 26 '22 at 08:36

Cristian-Florin Calina · Accepted Answer · 2022-06-26T08:35:46.100

In javascript, Array.pop() removes the last element. You need Array.shift() for removing the first one.

function permute_vector(row, p)

{

  if (p===0) // in js, checks are with == (loose) or === (strict).
             // = is for assignment and you were assigning p to 0 here.

  {return row}

   

  let q = new Array()

  for (i=0; i<4; i++)

  {

    q.push(row[i])

  }

 

for (i=0; i<p; i++)

  {

    q.push(q[0])

    q.shift() // shift instead of pop

  }

 

  for (i=0; i<4; i++)

  {

    row[i] = q[0]

    q.shift() // shift instead of pop

  }

return row

}

 

px = permute_vector([2,4,1,3], 1)

console.log("px is:", px)

}

score 0 · Answer 2 · answered Jun 26 '22 at 08:41

Two mistakes:

Comparisons need double or triple equal signs. In fact, you should prefer triple equal signs as you can read in this question/answer.
In python pop() accepts an index to remove a specific item. You're making use of that to remove the first item. In JavaScript Array.pop() merely is capable of removing the last item. Use Array.shift() to remove the first item.

Working code:

function permute_vector(row, p) {
  if (p === 0) {
    return row;
  }
  let q = new Array();
  for (i = 0; i < 4; i++) {
    q.push(row[i]);
  }
  for (i = 0; i < p; i++) {
    q.push(q[0]);

    q.shift();
  }
  for (i = 0; i < 4; i++) {
    row[i] = q[0];

    q.shift();
  }
  return row;
}

px = permute_vector([2, 4, 1, 3], 1);

console.log("px is:", px);

score 0 · Answer 3 · answered Jun 26 '22 at 09:38

Now when you've got it, you can just ditch the whole thing and replace it with

vec[n:] + vec[:n] (Python)

[...vec.slice(n), ...vec.slice(0, n)]  (JS)

Basic textbook algorithms are rarely useful in scripting languages, because they implement stuff like this out of the box.

Implementing pseudocode in JavaScript

3 Answers3