Implementing pseudocode in JavaScript

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!

Answer

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)

}