Converting large numbers from binary to decimal and back in JavaScript

后端 未结 4 1912
感动是毒
感动是毒 2020-11-29 11:54

I have a very large number represented as binary in JavaScript:

 var largeNumber = \'110100110101101000010100111110100101110111110000100101110001111100111110         


        
4条回答
  •  迷失自我
    2020-11-29 12:32

    As noted in Andrew L.'s answer, and by several commenters, your largeNumber exceeds what JavaScript can represent as an integer in an ordinary number without loss of precision—which is 9.007199254740991e+15.

    If you want to work with larger integers, you will need a BigInt library or other special-purpose code.

    Below is some code demonstrating how to convert arbitrarily large positive integers between different base representations, showing that the exact decimal representation of your largeNumber is

    15 798 770 299 367 407 029 725 345 423 297 491 683 306 908 462 684 165 669 735 033 278 996 876 231 474 309 788 453 071 122 111 686 268 816 862 247 538 905 966 252 886 886 438 931 450 432 740 640 141 331 094 589 505 960 171 298 398 097 197 475 262 433 234 991 526 525

    function parseBigInt(bigint, base) {
      //convert bigint string to array of digit values
      for (var values = [], i = 0; i < bigint.length; i++) {
        values[i] = parseInt(bigint.charAt(i), base);
      }
      return values;
    }
    
    function formatBigInt(values, base) {
      //convert array of digit values to bigint string
      for (var bigint = '', i = 0; i < values.length; i++) {
        bigint += values[i].toString(base);
      }
      return bigint;
    }
    
    function convertBase(bigint, inputBase, outputBase) {
      //takes a bigint string and converts to different base
      var inputValues = parseBigInt(bigint, inputBase),
        outputValues = [], //output array, little-endian/lsd order
        remainder,
        len = inputValues.length,
        pos = 0,
        i;
      while (pos < len) { //while digits left in input array
        remainder = 0; //set remainder to 0
        for (i = pos; i < len; i++) {
          //long integer division of input values divided by output base
          //remainder is added to output array
          remainder = inputValues[i] + remainder * inputBase;
          inputValues[i] = Math.floor(remainder / outputBase);
          remainder -= inputValues[i] * outputBase;
          if (inputValues[i] == 0 && i == pos) {
            pos++;
          }
        }
        outputValues.push(remainder);
      }
      outputValues.reverse(); //transform to big-endian/msd order
      return formatBigInt(outputValues, outputBase);
    }
    
    var largeNumber =
      '1101001101011010000101001111101001011101' + 
      '1111000010010111000111110011111011111000' +
      '0011000001100000110000011001110101001110' +
      '1010111010001000101101010111001111000001' +
      '1000001100000110000011001001100000110000' +
      '0110000011000001100001110000111000001100' +
      '0001100000110000011000001100001010110001' +
      '1001110101101001100110100100000110000011' +
      '0000011000001100010011010111101100100010' +
      '1101000110101101010001100100111000111001' +
      '0100111011011111010000110001110010101010' +
      '0011110100100001011000010000011000010110' +
      '0001101111100001111000111011111001111111' +
      '1000100011110110101000101100000110000011' +
      '0000011000001100000110100111010101101011' +
      '0100111110100101001011110101100001110110' +
      '0110010011001001111101';
    
    //convert largeNumber from base 2 to base 10
    var largeIntDecimal = convertBase(largeNumber, 2, 10);
    
    
    function groupDigits(bigint){//3-digit grouping
      return bigint.replace(/(\d)(?=(\d{3})+$)/g, "$1 ");
    }
    
    //show decimal result in console:
    console.log(groupDigits(largeIntDecimal));
    
    //converting back to base 2:
    var restoredOriginal = convertBase(largeIntDecimal, 10, 2);
    
    //check that it matches the original:
    console.log(restoredOriginal === largeNumber);

提交回复
热议问题