Vjeux

Daniel Baulig, a co-worker at Facebook, told me a little trick related to jDataView function to convert from a uint8 to a int8 in Javascript.

Here’s the version I had:

function getInt8() {
  var b = this.getUint8();
  if (b > Math.pow(2, 7) - 1) {
    return b - Math.pow(2, 8);
  }
  return b;
}

Compare it to his version:

< <function getInt8() {
  return this.getUint8() << 24 >> 24;
}

I was really confused because it seems like it’s doing a no-op. Here’s the full explanation of why the two versions are working.

How it works?

The following table (borrowed from Wikipedia) shows how various 8 bits values are in represented with bits and how they are interpreted in unsigned and signed (using two-complement rule).

.uint8table {margin: 0 auto; border-collapse: collapse; }
.uint8table td, .uint8table th { padding: 2px 5px; border: 1px solid #ccc; text-align: center; }
.uint8table b { background-color: #FFE0ED; font-weight: normal; }
.uint8table i { color: purple; font-style: normal; }
.uint8table .sep td { border-bottom: 2px solid #bbb; }
#hidefirsttwo span:nth-child(1), #hidefirsttwo span:nth-child(2) { display: none; }
#hidefirsttwo span:nth-child(3) { margin-left: -7px; }

Javascript doesn’t natively have a 8 bit integer type, it only has a 32 bits one. When you put a 8 bit integer into a 32 bits one, Javascript is going to fill the remaining bits on the left with zeros as the following table shows.

Unfortunately, this doesn’t properly handle negative numbers. Because we use two-complement, we’ve got to fill all the bits with 1 for negative numbers in order to have the same number in a signed 32 bits representation.

So basically, we’ve got to fill the 24 remaining bits on the left with the same first bit we have: 0 for positive numbers and 1 for negative numbers.

This is when the trick comes into place. In javascript, there’s a binary operator: >> >_(Sign-propagating_right_shift)”>Sign-propagating right shift that moves all the bits to the right and fills the missing bits with the first bit.

So all we have to do is to put our 8 good digits to the far left using << and then use the previous trick to fill the bits with the proper ones 🙂