What do “>>” and “<<” mean in Javascript?

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清酒与你
清酒与你 2020-11-29 13:39

I have a piece of Javascript code I\'m trying to understand

// read big-endian (network byte order) 32-bit float
readFloat32 = function(data, offset) {
    v         


        
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  • 2020-11-29 13:41

    These are the shift right (with sign) and shift left operators.

    Essentially, these operators are used to manipulate values at BIT-level.
    They are typically used along with the the & (bitwise AND) and | (bitwise OR) operators and in association with masks values such as the 0x7F and similar immediate values found the question's snippet.
    The snippet in question uses these operators to "parse" the three components of a 32 bits float value (sign, exponent and fraction).

    For example, in the question's snippet:
    1 - (2*(b1 >> 7)) produces the integer value 1 or -1 depending if the bit 7 (the 8th bit from the right) in the b1 variable is zero or one respectively.
    This idiom can be explained as follow.

    • at the start, b1, expressed as bits is 0000000000000000abcdefgh
      note how all the bits on the left are zeros, this comes from the
      b1 = data.charCodeAt(offset) & 0xFF assignement a few lines above, which essentially zero-ed all the bits in b1 except for the rightmot 8 bits (0xFF mask).
      a, b, c... thru h represent unknown boolean values either 0 or 1.
      We are interested in testing the value of a.
    • b1 >> 7 shifts this value to the right by 7 bits, leaving
      b1 as 00000000000000000000000a which, read as an integer will have value 1 or 0
    • this 1 or 0 integer value is then multiplied by 2
      it is then either 2 or 0, respectively.
    • this value is then substracted from 1, leaving either -1 or 1.

    Although useful to illustrate the way the bit-operators work, the above idiom could be replaced by something which tests the bit 7 more directly and assigns the sign variable more explicitly. Furthermore this approach does not require the initial masking of the leftmost bits in b1:

    var sign
    if (b1 & 0x80)   // test bit 7  (0x80 is  [00000000]10000000)
      sign = -1;
    else
      sign = 1;
    
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  • 2020-11-29 13:48

    It's the bitshifting operator. See here for more details.

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  • 2020-11-29 13:51

    Right and Left and shift operators.

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  • 2020-11-29 13:53

    You can read about the operators here: https://developer.mozilla.org/en/JavaScript/Reference/operators/bitwise_operators

    They are bit shifts and also occur in languages other than JS.

    Example: 5 >> 1 = 2

    binary: 0101 shifting one position = 0010

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  • 2020-11-29 13:55

    They are bitshift operators. Numbers in the computer are represented in binary. Shifting left is equivalent to multiplying by 2 and shifting right is equivalent to dividing by 2.

    For example, the number 8 is 1000 in binary. Shift left << by 3 would yield 1000000 which is 64. Shift right by 2 would yield 10 which is 2.

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  • 2020-11-29 13:56

    shift left and shift right operators. If you have a number it will shift its bits to left or right.

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