JavaScript implementation of Math.pow

Question

I implemented Math.pow using a log(n) solution just like this article on geeksforgeeks

http://www.geeksforgeeks.org/write-a-c-program-to-calculate-powxn/

However, I'm finding that the function does not exit its base case as I had intended. This program seems like it works in C but not JS.

Therefore, I am concluding that there is something about C that I am assuming works as well in JavaScript.

What am I missing in my JavaScript implementation?

Be forewarned: the codesnippet as it is will have a max call stack exceeded error

var myPow = function(x, n) {
  var res = 1
  var temp;
  if (n === 0) {
    return 1;
  }
  temp = myPow(x, n / 2)
  if (n % 2 === 0) {
    return temp * temp
  } else {
    return x * temp * temp
  }
};

console.log(myPow(2,3));

@Anthony :hope to find it useful ? http://stackoverflow.com/a/38666376/747579 — Abdennour TOUMI, Jul 29 '16 at 19:52
Your recursive call isn't in tail position. Here's a tail recursive ES2015 solution: `const power = (base, exp, acc = 1) => exp === 0 ? acc : power(base, exp - 1, base * acc)`. — , Jul 29 '16 at 19:56
Take a look at iterative (non recursive) approaches on integer math for some additional ideas: [Power by squaring for negative exponents](http://stackoverflow.com/a/30962495/2521214) — Spektre, Jul 30 '16 at 06:45

score 6 · Accepted Answer · edited Jun 20 '20 at 09:12

6

Brief :

use parseInt or Math.floor to have y/2 as integer, unleness you will not reach 0 which is the stopper of recursion .

Details

if you want to transalte [C Algo]:

int power(int x, unsigned int y)
{
    if( y == 0)
        return 1;
    else if (y%2 == 0)
        return power(x, y/2)*power(x, y/2);
    else
        return x*power(x, y/2)*power(x, y/2);
 
}

To [JS Algo] , you will have :

function power(x,y){
     if(y===0){return 1}
     else if (y%2 ===0){
         return power(x,parseInt(y/2))*power(x,parseInt(y/2))
     }else{
          return x*power(x,parseInt(y/2))*power(x,parseInt(y/2))
     }

}

DEMO :

    function power(x,y){
         if(y===0){return 1}
         else if (y%2 ===0){
             return power(x,parseInt(y/2))*power(x,parseInt(y/2))
         }else{
              return x*power(x,parseInt(y/2))*power(x,parseInt(y/2))
         }
    
    }


console.log(power(3,2))

edited Jun 20 '20 at 09:12

Community

1
1

answered Jul 29 '16 at 19:41

Abdennour TOUMI

87,526
38
249
254

While this is generally safe, it's standard best practice to always provide the radix for parseInt: `parseInt(y/2, 10)` – rgthree Jul 29 '16 at 19:43
3

`parseInt()` parses a string and returns an integer. You should be using [`Math.floor()`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/floor) instead to truncate the decimal places – Andreas Jul 29 '16 at 19:44
@Andreas &@rgthree : THank u ,, any way , the idea is to reach `0` at the end of recursion to Avoid `Infinity` – Abdennour TOUMI Jul 29 '16 at 19:48
3

Note that bitwise or `|` with 0 also works: `2.3434 | 0 === 2` – Jared Smith Jul 29 '16 at 20:35
Nice job! any idea on how you would have this work with negative exponentiation as well? – Anthony Chung Jul 30 '16 at 17:03
You shouldn't misuse `parseInt` to perform a `floor` operation. That wouldn't work on larger numbers. E.g., `parseInt(1000000000000000000000000/2)` is `5`. – Boann Jul 31 '16 at 11:38
To make it more simpler, power(x, parseInt(y/ 2)) can be stored and accessed instead of calling it again. Still, a great example ! – Arunkumar Srisailapathi Nov 29 '17 at 17:53
1

@AnthonyChung . – Abdennour TOUMI Nov 29 '17 at 18:43
Is there any way to implement this using javascript reducers? – Gal Bracha Feb 28 '18 at 03:11
The question is requesting to translate the code from programming language to another. Otherwise, I totally agree with you to use reducers. – Abdennour TOUMI Feb 28 '18 at 08:52
ok, so from when 3^-2 is 9. – Ace Jul 20 '22 at 04:20

Shanto Islam · Answer 2 · 2022-10-03T08:33:01.810

Try this out

It will give you the same result of JavaScript build in method ( Math.pi(x, y)) but the only problem is you can't use Power as decimal number.

Math.my_pow = (x, y) => {
  if (typeof x != "number" || typeof y != "number")
    throw "(x) and (y) should only be number";

  if (y == 0) return 1;
  if (x == 0 && y > 0 ) return 0;

  const base = x;
  var value = base;
  var pow = y;
  if (y < 0) pow = y * -1;

  for (var i = 1; i < pow; i++) {
    value *= base;
  }

  if (y < 0) return 1 / value;
  return value;
};

try {
  console.log( Math.my_pow(0, -3) );
  console.log( Math.pow(0, -2) );

  console.log( Math.my_pow(-5, -3) );
  console.log( Math.pow(-5, -3) );

  console.log( Math.my_pow(8, -7) );
  console.log( Math.pow(8, -7)) ;
} catch (err) {
  console.log(err);
}

definitely more readable than the accepted recursive solution. — EugenSunic, Nov 11 '22 at 12:59

score 0 · Answer 3 · answered Nov 29 '17 at 17:57

Modifying a little bit from @Abdennour's answer

function power(x, y) {
    if (y === 0) {
      return 1;
    }
    var yBy2 = y / 2;
    var pow = power(x, parseInt( yBy2, 10) );
    if (y % 2 === 0) {
      return pow * pow;
    } else {
      return x * pow * pow;
    }
}

score -1 · Answer 4 · answered Mar 11 '19 at 06:25

-1

* Simple Logic *

function pow (p,y){
    if(p > 0 && y === 0){
        return 1;
    }
    x = p;
    while(y > 1){
        x = p*x;
        y--;
    }
    return x;
}


console.log(pow(0,0)+'\n'); // 0
console.log(pow(2,4)+'\n'); // 16
console.log(pow(2,0)+'\n'); // 1
console.log(pow(0,2)+'\n'); // 0

answered Mar 11 '19 at 06:25

Manish Goswami

442
4
10

We cannot handle negative power in this solution. – Awais Ibrar Jan 02 '22 at 18:22

JavaScript implementation of Math.pow

4 Answers4

Brief :

Details

DEMO :

Try this out