JavaScript Conversion from Binary to Decimal

Question

JavaScript Conversion from Binary to Decimal

This code is meant to convert a binary string ("100101") to its decimal equivalent. I am struggling to identify the issue causing it to not work properly. Any assistance would be greatly appreciated.

function BinaryConverter(str) { 
 var num=str.split("");
 var powers=[];
 var sum=0;
  for(var i=0;i<num.length;i++){
   powers.push(i);
}
 for(var i=powers.length-1;i>=0;i--){
  for(var j=0;j<num.length;i++){
   sum+=Math.pow(2,i)*num[j];
  }
 }
 return sum;
};

Below is my revised code. For an input of "011", it is intended to compute (2^2*0 +2^1*1 +2^0*1) which should be 3, but currently, it outputs 14. Can anyone pinpoint where I've made a mistake?

function BinaryConverter(str) { 
 var num=str.split("");
 var powers=[];
 var sum=0;
 for(var i=0;i<num.length;i++){
   powers.push(i);
 }
 for(var i=powers.length-1;i>=0;i--){
  for(var j=0;j<num.length;j++){
   sum+=Math.pow(2,i)*num[j];
  }
 }
 return sum;
};

javascript math binary numbers

Answer 1

Answer №1

An issue arises with the two nested for loops in your code. The first loop subtracts an i, while the second loop adds an i, resulting in an endless loop.

Your revised code should look like this:

function BinaryConverter(str) { 
    var num = str.split("");
    var powers = [];
    var sum = 0;
    var numlength = num.length;

    for (var i = 0; i < num.length; i++) {
        powers.push(i);
    }

    for (var i = powers.length - 1; i >= 0; i--) {
        sum += Math.pow(2, i) * num[numlength - i - 1];
    }

    return sum;
};

It seems that eliminating the nested for loop would be a better approach.

Answer 2

An issue arises with the two nested for loops in your code. The first loop subtracts an i, while the second loop adds an i, resulting in an endless loop.

Your revised code should look like this:

function BinaryConverter(str) { 
    var num = str.split("");
    var powers = [];
    var sum = 0;
    var numlength = num.length;

    for (var i = 0; i < num.length; i++) {
        powers.push(i);
    }

    for (var i = powers.length - 1; i >= 0; i--) {
        sum += Math.pow(2, i) * num[numlength - i - 1];
    }

    return sum;
};

It seems that eliminating the nested for loop would be a better approach.

Answer 3

Answer №2

Instead of using parseInt() to convert binary strings, you can implement a more efficient solution without relying on Math.pow() for each digit:

function parseBinary(str) {
  var i, value = 0;
  for (i = 0; i < str.length; ++i)
    value = value * 2 + +str[i];
  return value;
}

It's important to note that this function does not validate input strings for invalid binary characters.

Answer 4

Instead of using parseInt() to convert binary strings, you can implement a more efficient solution without relying on Math.pow() for each digit:

function parseBinary(str) {
  var i, value = 0;
  for (i = 0; i < str.length; ++i)
    value = value * 2 + +str[i];
  return value;
}

It's important to note that this function does not validate input strings for invalid binary characters.

Answer 5

Answer №3

ace040686 made a minor adjustment in his answer, where he reversed the order of pow(2,i) and num[len-1-i]. Other than that, the answer is correct. Additionally, adding 0..str.length-1 to powers is unnecessary as these are implicit indices.

function convertNaive(str) {
  var num = str.split("");
  var len = num.length;
  var sum = 0;
  for(var i = len - 1; i >= 0; --i)
    sum += Math.pow(2, len - 1 - i) * num[i];
  return sum;
}

You can optimize this further to remove the unnecessary array and reduce the usage of Math.pow:

function convertImproved(str) {
  var len = str.length;
  var sum = 0;
  for(var i = 0, fac = 1; i < len; ++i, fac *= 2)
    sum += fac * str[len - 1 - i];
  return sum;
}

Give it a try yourself:

var input = "100101";
var logNode = document.getElementById("log");

function log(line) {
  var text = document.createTextNode(line);
  var node = document.createElement("p");
  node.appendChild(text);
  logNode.appendChild(node);
}

function convertNaive(str) {
  var num = str.split("");
  var len = num.length;
  var sum = 0;
  for(var i = len - 1; i >= 0; --i)
    sum += Math.pow(2, len - 1 - i) * num[i];
  return sum;
}

function convertImproved(str) {
  var len = str.length;
  var sum = 0;
  for(var i = 0, fac = 1; i < len; ++i, fac *= 2)
    sum += fac * str[len - 1 - i];
  return sum;
}

log("input: " + input);
log("parseInt(input, 2): " + parseInt(input, 2));
log("convertNaive(input): " + convertNaive(input));
log("convertImproved(input): " + convertImproved(input));

<div id="log" />

Answer 6

ace040686 made a minor adjustment in his answer, where he reversed the order of pow(2,i) and num[len-1-i]. Other than that, the answer is correct. Additionally, adding 0..str.length-1 to powers is unnecessary as these are implicit indices.

function convertNaive(str) {
  var num = str.split("");
  var len = num.length;
  var sum = 0;
  for(var i = len - 1; i >= 0; --i)
    sum += Math.pow(2, len - 1 - i) * num[i];
  return sum;
}

You can optimize this further to remove the unnecessary array and reduce the usage of Math.pow:

function convertImproved(str) {
  var len = str.length;
  var sum = 0;
  for(var i = 0, fac = 1; i < len; ++i, fac *= 2)
    sum += fac * str[len - 1 - i];
  return sum;
}

Give it a try yourself:

var input = "100101";
var logNode = document.getElementById("log");

function log(line) {
  var text = document.createTextNode(line);
  var node = document.createElement("p");
  node.appendChild(text);
  logNode.appendChild(node);
}

function convertNaive(str) {
  var num = str.split("");
  var len = num.length;
  var sum = 0;
  for(var i = len - 1; i >= 0; --i)
    sum += Math.pow(2, len - 1 - i) * num[i];
  return sum;
}

function convertImproved(str) {
  var len = str.length;
  var sum = 0;
  for(var i = 0, fac = 1; i < len; ++i, fac *= 2)
    sum += fac * str[len - 1 - i];
  return sum;
}

log("input: " + input);
log("parseInt(input, 2): " + parseInt(input, 2));
log("convertNaive(input): " + convertNaive(input));
log("convertImproved(input): " + convertImproved(input));

<div id="log" />

Answer 7

Answer №4

Check out this straightforward Javascript code for converting binary numbers to decimal.

run();

function run() {
    let binaryInput = 10000100111;
    let decimalOutput = convertBinaryToDecimal(binaryInput);
    console.log(decimalOutput);
}

function convertBinaryToDecimal(input) {
    let binaryString = input.toString();
    let result = 0;
    let exponent = 1;
    let currentBit = 0;
    for (let i = binaryString.length - 1; i >= 0; i--) {
        currentBit = parseInt(binaryString[i]);
        currentBit *= exponent;
        result += currentBit;
        exponent *= 2;
    }
    return result;
}

Answer 8

Check out this straightforward Javascript code for converting binary numbers to decimal.

run();

function run() {
    let binaryInput = 10000100111;
    let decimalOutput = convertBinaryToDecimal(binaryInput);
    console.log(decimalOutput);
}

function convertBinaryToDecimal(input) {
    let binaryString = input.toString();
    let result = 0;
    let exponent = 1;
    let currentBit = 0;
    for (let i = binaryString.length - 1; i >= 0; i--) {
        currentBit = parseInt(binaryString[i]);
        currentBit *= exponent;
        result += currentBit;
        exponent *= 2;
    }
    return result;
}

JavaScript Conversion from Binary to Decimal

Answer №1

Answer №2

Answer №3

Answer №4

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