如何将二进制数字转换为带符号的十进制数字 [英] How to convert a binary number into a Signed decimal number

问题描述

我有一个类似这样的二进制数.var data ="1100110010". 我正在尝试将此转换为Javascript中适当的带符号十进制数.

I have a binary number something like this.. var data = "1100110010". I am trying to convert this into appropriate signed decimal number in Javascript.

我使用的步骤如下

1)var data = "1100110010"; var bin2dec = parseInt(data,2).toString(10);

1) var data = "1100110010"; var bin2dec = parseInt(data,2).toString(10);

bin2dec给出转换后的Decimal数字为"818".

The bin2dec gives the converted Decimal number as "818" .

但是我也想访问二进制数字的带符号十进制.

But I want to access the signed decimal of the binary number too.

此二进制代码的带符号十进制数字是"-206".

The signed decimal number for this binary code is "-206".

如何从给定的二进制数中访问带符号和无符号十进制值.请告诉我 .任何想法都会有所帮助

How can I access both signed and unsigned decimal Value from a given Binary Number. Please let me know . Any ideas would help

推荐答案

使用一些移位技巧

function uintToInt(uint, nbit) {
    nbit = +nbit || 32;
    if (nbit > 32) throw new RangeError('uintToInt only supports ints up to 32 bits');
    uint <<= 32 - nbit;
    uint >>= 32 - nbit;
    return uint;
}

uintToInt(818, 10); // -206

• 为什么818?因为这是您的二进制字符串

  • Why 818? Because this is the uint value of your binary string

    parseInt('1100110010', 2); // 818
    

  • 为什么10?因为您的带符号int由二进制字符串

  • Why 10? Because your signed int is represented by 10 bits in your binary string

    '1100110010'.length; // 10
    

  请注意，对于正数，您不能仅使用nbit = str.length;，因为字符串可能未填充0，因此您需要知道您实际使用了多少位

  Please note that for positive numbers, you can't just take nbit = str.length; as the string may not be 0-padded, you'll need to know how many bits you're actually using

  您可能还想为uint > 4294967295

  出于完整性考虑，

  function intToUint(int, nbit) {
      var u = new Uint32Array(1);
      nbit = +nbit || 32;
      if (nbit > 32) throw new RangeError('intToUint only supports ints up to 32 bits');
      u[0] = int;
      if (nbit < 32) { // don't accidentally sign again
          int = Math.pow(2, nbit) - 1;
          return u[0] & int;
      } else {
          return u[0];
      }
  }
  
  intToUint(-206, 10); // 818
  

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