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I am working on parsing some binary files, I have them opened and in an ArrayBuffer.

In the particular file structure I am reading, there are a number of bits which are boolean and I can check whether they are checked with:

(flag & 1) != 0; // bit 0 
(flag & 2) != 0; // bit 1 
(flag & 4) != 0; // bit 2

etc.

However, I am having trouble getting the values of the bits followed. They span over multiple bits (for example bits 4-6) and consist of an integer value from 0-7.

How are multiple bits read like that? I understand that this isn't as much of a JavaScript question than that of how bits and bitwise operators work.

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2 Answers 2

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Assuming you want 4-6 bits from a byte like this:

76543210
 ^^^

You would construct a bit mask like this:

0x70

which means:

01110000

And then you would & that with the number and shift to right 4 times:

( byte & 0x70 ) >> 4
//Number between 0-7
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9 Comments

In javascript, I would still always recommend shifting with >>> instead of >>. (In C, cast to unsigned before shifting).
@selbie I didn't mention this in my question, but if I've read the byte as Uint, it will be unsigned already?
the & already drops out potential sign bit, which is in position 31
There is no such thing as an "unsigned integer" in javascript. There is a "number" type, and that is it. So Mike, I'm not sure how you are reading a UINT in javascript. @Esailija - where is the behavior of & operator making the value unsigned referenced? Or is it just a side effect of javascript not really having an upper bound on number types?
@selbie, all numbers are converted to signed 32 bit integers (with exception >>>, converts to unsigned 32) in javascript before the bitwise operation. So if it was a signed 8 bit integer -127, it would be 0xFFFFFF81before the & operation, and 0x0000000 after in that case. In any case, the sign bit is not persisted through the AND.
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Assuming the least significant bit is at position "0", and you want the 3-bit integer between bit positions 4-6.

var value = (flag >>> 4) & 0x0007;

In other words, right shift "flag" 4 bits to the right, such that bits 4-6 get shifted into positions 0-2. Then mask off just the last three bits (binary 111 = decimal 7).

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I updated my answer below to use the javascript "unsigned shift" operator (>>>) instead of the more usual >> operator. Always a good idea in any language to make sure you are doing bit-shifting on unsigned types, because you might get a side-effect if the value being shifted is negative (1's imported on the left side instead of 0's).

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