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How to clear all bits above a certain bit position in Javascript

I have a bit manipulation question in Javascript.
I have a number that I need to clear some upper bits from. Which exact bits need to be cleared is determined programmatically and as a result of some calculation, I have two pieces of information, a bit position and a mask for that particular bit position. What I want to do is to clear all the bits above that bit position (not include that bit) in my value.
So, if my value is:
0b11111111
And, a calculation has yielded bit position 3 and a mask for that bit position:
0b00000100
I want to mask out or clear the bits above that bit position 3 in my value and end up with:
0b00000111
As a reminder, the bit position that I want to clear above is programmatically determined and could be any bit position so I can't just code a pre-defined mask. How would I do that? Programming environment is Javascript in node.js.

value & ((mask << 1) - 1)
In the example given, 100 leftshifted by 1 will be 1000 (8) minus one will be 111 (7). value & 111 will clear all bits except the last three. Might not work with more than 30 bits. If 31 bits are needed, one could also use a very similar formula:
value & (mask * 2 - 1)
The logic is pretty much the same, but as the multiplication gets performed on a 52 bit mantissa, you get one bit more.

(Novice Programmer) mod(3^146, 293) among others returning the same incorrect values in Matlab and JS

First note that mod(3^146,293)=292. For some reason, inputting mod(3^146,293) in Matlab returns 275. Inputting Math.pow(3,146) % 293 in JS returns 275. This same error occurs (as far as I can tell) every time. This leads me to believe I am missing something obvious but cannot seem to tell what.
Any help is much appreciated.

As discussed in the answers to this related question, MATLAB uses double-precision floating point numbers by default, which have limits on their resolution (i.e. the floating point relative accuracy, eps). For example:
>> a = 3^146
a =
4.567759074507741e+69
>> eps(a)
ans =
7.662477704329444e+53
In this case, 3146 is on the order of 1069 and the relative accuracy is on the order of 1053. With only 16 digits of precision, a double can't store the exact integer representation of an arbitrary 70 digit integer.
An alternative in MATLAB is to use the Symbolic Toolbox to create symbolic numbers with a greater resolution. This gives you the answer you expect:
>> a = sym('3^146')
a =
4567759074507740406477787437675267212178680251724974985372646979033929
>> mod(a, 293)
ans =
292

Math.pow(3, 146) is is larger than the constant Number.MAX_SAFE_INTEGER in JavaScript which represents the upper limit of numbers that can be represented without losing any accuracy. Therefore JavaScript cannot accurately represent Math.pow(3, 146) within the 64 bit limit.
MatLab also has limits on its integer size but can represent a large number with the Symbolic Math Toolbox.
There are also algorithms that you can implement to accomplish this without overflowing.

Reassembling negative Python marshal int's into Javascript numbers

I'm writing a client-side Python bytecode interpreter in Javascript (specifically Typescript) for a class project. Parsing the bytecode was going fine until I tried out a negative number.
In Python, marshal.dumps(2) gives 'i\x02\x00\x00\x00' and marshal.dumps(-2) gives 'i\xfe\xff\xff\xff'. This makes sense as Python represents integers using two's complement with at least 32 bits of precision.
In my Typescript code, I use the equivalent of Node.js's Buffer class (via a library called BrowserFS, instead of ArrayBuffers and etc.) to read the data. When I see the character 'i' (i.e. buffer.readUInt8(offset) == 105, signalling that the next thing is an int), I then call readInt32LE on the next offset to read a little-endian signed long (4 bytes). This works fine for positive numbers but not for negative numbers: for 1 I get '1', but for '-1' I get something like '-272777233'.
I guess that Javascript represents numbers in 64-bit (floating point?). So, it seems like the following should work:
var longval = buffer.readInt32LE(offset); // reads a 4-byte long, gives -272777233
var low32Bits = longval & 0xffff0000; //take the little endian 'most significant' 32 bits
var newval = ~low32Bits + 1; //invert the bits and add 1 to negate the original value
//but now newval = 272826368 instead of -2
I've tried a lot of different things and I've been stuck on this for days. I can't figure out how to recover the original value of the Python integer from the binary marshal string using Javascript/Typescript. Also I think I deeply misunderstand how bits work. Any thoughts would be appreciated here.
Some more specific questions might be:
Why would buffer.readInt32LE work for positive ints but not negative?
Am I using the correct method to get the 'most significant' or 'lowest' 32 bits (i.e. does & 0xffff0000 work how I think it does?)
Separate but related: in an actual 'long' number (i.e. longer than '-2'), I think there is a sign bit and a magnitude, and I think this information is stored in the 'highest' 2 bits of the number (i.e. at number & 0x000000ff?) -- is this the correct way of thinking about this?

The sequence ef bf bd is the UTF-8 sequence for the "Unicode replacement character", which Unicode encoders use to represent invalid encodings.
It sounds like whatever method you're using to download the data is getting accidentally run through a UTF-8 decoder and corrupting the raw datastream. Be sure you're using blob instead of text, or whatever the equivalent is for the way you're downloading the bytecode.
This got messed up only for negative values because positive values are within the normal mapping space of UTF-8 and thus get translated 1:1 from the original byte stream.

Hash 32bit int to 16bit int?

What are some simple ways to hash a 32-bit integer (e.g. IP address, e.g. Unix time_t, etc.) down to a 16-bit integer?
E.g. hash_32b_to_16b(0x12345678) might return 0xABCD.
Let's start with this as a horrible but functional example solution:
function hash_32b_to_16b(val32b) {
return val32b % 0xffff;
}
Question is specifically about JavaScript, but feel free to add any language-neutral solutions, preferably without using library functions.
The context for this question is generating unique IDs (e.g. a 64-bit ID might be composed of several 16-bit hashes of various 32-bit values). Avoiding collisions is important.
Simple = good. Wacky+obfuscated = amusing.

The key to maximizing the preservation of entropy of some original 32-bit 'signal' is to ensure that each of the 32 input bits has an independent and equal ability to alter the value of the 16-bit output word.
Since the OP is requesting a bit-size which is exactly half of the original, the simplest way to satisfy this criteria is to xor the upper and lower halves, as others have mentioned. Using xor is optimal because—as is obvious by the definition of xor—independently flipping any one of the 32 input bits is guaranteed to change the value of the 16-bit output.
The problem becomes more interesting when you need further reduction beyond just half-the-size, say from a 32-bit input to, let's say, a 2-bit output. Remember, the goal is to preserve as much entropy from the source as possible, so solutions which involve naively masking off the two lowest bits with (i & 3) are generally heading in the wrong direction; doing that guarantees that there's no way for any bits except the unmasked bits to affect the result, and that generally means there's an arbitrary, possibly valuable part of the runtime signal which is being summarily discarded without principle.
Following from the earlier paragraph, you could of course iterate with xor three additional times to produce a 2-bit output with the desired property of being equally-influenced by each/any of the input bits. That solution is still optimally correct of course, but involves looping or multiple unrolled operations which, as it turns out, aren't necessary!
Fortunately, there is a nice technique of only two operations which gives the same optimal result for this situation. As with xor, it not only ensures that, for any given 32-bit value, twiddling any input bit will result in a change to the 2-bit output, but also that, given a uniform distribution of input values, the distribution of 2-bit output values will also be perfectly uniform. In the current example, the method divides the 4,294,967,296 possible input values into exactly 1,073,741,824 each of the four possible 2-bit hash results { 0, 1, 2, 3 }.
The method I mention here uses specific magic values that I discovered via exhaustive search, and which don't seem to be discussed very much elsewhere on the internet, at least for the particular use under discussion here (i.e., ensuring a uniform hash distribution that's maximally entropy-preserving). Curiously, according to this same exhaustive search, the magic values are in fact unique, meaning that for each of target bit-widths { 16, 8, 4, 2 }, the magic value I show below is the only value that, when used as I show here, satisfies the perfect hashing criteria outlined above.
Without further ado, the unique and mathematically optimal procedure for hashing 32-bits to n = { 16, 8, 4, 2 } is to multiply by the magic value corresponding to n (unsigned, discarding overflow), and then take the n highest bits of the result. To isolate those result bits as a hash value in the range [0 ... (2ⁿ - 1)], simply right-shift (unsigned!) the multiplication result by 32 - n bits.
The "magic" values, and C-like expression syntax are as follows:
Method
Maximum-entropy-preserving hash for reducing 32 bits to. . .
Target Bits Multiplier Right Shift Expression [1, 2]
----------- ------------ ----------- -----------------------
16 0x80008001 16 (i * 0x80008001) >> 16
8 0x80808081 24 (i * 0x80808081) >> 24
4 0x88888889 28 (i * 0x88888889) >> 28
2 0xAAAAAAAB 30 (i * 0xAAAAAAAB) >> 30
Maximum-entropy-preserving hash for reducing 64 bits to. . .
Target Bits Multiplier Right Shift Expression [1, 2]
----------- ------------------ ----------- -------------------------------
32 0x8000000080000001 32 (i * 0x8000000080000001) >> 32
16 0x8000800080008001 48 (i * 0x8000800080008001) >> 48
8 0x8080808080808081 56 (i * 0x8080808080808081) >> 56
4 0x8888888888888889 60 (i * 0x8888888888888889) >> 60
2 0xAAAAAAAAAAAAAAAB 62 (i * 0xAAAAAAAAAAAAAAAB) >> 62
Notes:
Use unsigned multiply and discard any overflow (64-bit multiply is not needed).
If isolating the result using right-shift (as shown), be sure to use an unsigned shift operation.
Further discussion
I find this all this quite cool. In practical terms, the key information-theoretical requirement is the guar­antee that, for any m-bit input value and its corresponding n-bit hash value result, flipping any one of the m source bits always causes some change in the n-bit result value. Now al­though there are 2ⁿ possible result values in total, one of them is already "in-use" (by the result itself) since "switching" to that one from any other result would be no change at all. This leaves 2ⁿ - 1 result values that are eligible to be used by the entire set of m input values flipped by a single bit.
Let's consider an example; in fact, to show how this technique might seem to border on spooky or downright magical, we'll consider the more extreme case where m = 64 and n = 2. With 2 output bits there are four possible result values, { 0, 1, 2, 3 }. Assuming an arbitrary 64-bit input value 0x7521d9318fbdf523, we obtain its 2-bit hash value of 1:
(0x7521d9318fbdf523 * 0xAAAAAAAAAAAAAAAB) >> 62 // result --> '1'
So the result is 1 and the claim is that no value in the set of 64 values where a single-bit of 0x7521d9318fbdf523 is toggled may have that same result value. That is, none of those 64 other results can use value 1 and all must instead use either 0, 2, or 3. So in this example it seems like every one of the 2⁶⁴ input values—to the exclusion of 64 other input values—will selfishly hog one-quarter of the output space for itself. When you consider the sheer magnitude of these interacting constraints, can a simultaneously satisfying solution overall even exist?
Well sure enough, to show that (exactly?) one does, here are the hash result values, listed in order, for inputs that flipping a single bit of 0x7521d9318fbdf523 (one at a time), from MSB (position 63) down to LSB (0).
3 2 0 3 3 3 3 3 3 0 0 0 3 0 3 3 0 3 3 3 0 0 3 3 3 0 0 3 3 0 3 3 // continued…
0 0 3 0 0 3 0 3 0 0 0 3 0 3 3 3 0 3 0 3 3 3 3 3 3 0 0 0 3 0 0 3 // notice: no '1' values
As you can see, there are no 1 values, which entails that every bit in the source "as-is" must be contributing to influence the result (or, if you prefer, the de facto state of each-and-every bit in 0x7521d9318fbdf523 is essential to keeping the entire overall result from being "not-1"). Because no matter what single-bit change you make to the 64-bit input, the 2-bit result value will no longer be 1.
Keep in mind that the "missing-value" table shown above was dumped from the analysis of just the one randomly-chosen example value 0x7521d9318fbdf523; every other possible input value has a similar table of its own, each one eerily missing its owner's actual result value while yet somehow being globally consistent across its set-membership. This property essentially corresponds to maximally preserving the available entropy during the (inherently lossy) bit-width reduction task.
So we see that every one of the 2⁶⁴ possible source values independently imposes, on exactly 64 other source values, the constraint of excluding one of the possible result values. What defies my intuition about this is that there are untold quadrillions of these 64-member sets, each of whose members also belongs to 63 other, seemingly unrelated bit-twiddling sets. Yet somehow despite this most confounding puzzle of interwoven constraints, it is nevertheless trivial to exploit the one (I surmise) resolution which simultaneously satisfies them all exactly.
All this seems related to something you may have noticed in the tables above: namely, I don't see any obvious way to extend the technique to the case of compressing down to a 1-bit result. In this case, there are only two possible result values { 0, 1 }, so if any/every given (e.g.) 64-bit input value still summarily excludes its own result from being the result for all 64 of its single-bit-flip neighbors, then that now essentially imposes the other, only remaining value on those 64. The math breakdown we see in the table seems to be signalling that a simultaneous result under such conditions is a bridge too far.
In other words, the special 'information-preserving' characteristic of xor (that is, its luxuriously reliable guarantee that, as opposed to and, or, etc., it c̲a̲n̲ and w̲i̲l̲l̲ always change a bit) not surprisingly exacts a certain cost, namely, a fiercely non-negotiable demand for a certain amount of elbow room—at least 2 bits—to work with.

I think this is the best you're going to get. You could compress the code to a single line but the var's are there for now as documentation:
function hash_32b_to_16b(val32b) {
var rightBits = val32b & 0xffff; // Left-most 16 bits
var leftBits = val32b & 0xffff0000; // Right-most 16 bits
leftBits = leftBits >>> 16; // Shift the left-most 16 bits to a 16-bit value
return rightBits ^ leftBits; // XOR the left-most and right-most bits
}
Given the parameters of the problem, the best solution would have each 16-bit hash correspond to exactly 2^16 32-bit numbers. It would also IMO hash sequential 32-bit numbers differently. Unless I'm missing something, I believe this solution does those two things.
I would argue that security cannot be a consideration in this problem, as the hashed value is just too few bits. I believe that the solution I gave provides even distribution of 32-bit numbers to 16-bit hashes

This depends on the nature of the integers.
If they can contain some bit-masks, or can differ by powers of two, then simple XORs will have high probability of collisions.
You can try something like (i>>16) ^ ((i&0xffff) * p) with p being a prime number.
Security-hashes like MD5 are all good, but they are obviously an overkill here. Anything more complex than CRC16 is overkill.

I would say just apply a standard hash like sha1 or md5 and then grab the last 16 bits of that.

Assuming that you expect the least significant bits to 'vary' the most, I think you're probably going to get a good enough distribution by just using the lower 16-bits of the value as a hash.
If the numbers you're going to hash won't have that kind of distribution, then the additional step of xor-ing in the upper 16 bits might be helpful.
Of course this suggestion is if you're intending to use the hash merely for some sort of lookup/storage scheme and aren't looking for the crypto-related properties of non-guessability and non-reversability (which the xor-ing suggestions don't really buy you either).

Something simple like this....
function hash_32b_to_16b(val32b) {
var h = hmac(secretKey, sha512);
var v = val32b;
for(var i = 0; i < 4096; ++i)
v = h(v);
return v % 0xffff;
}

2.9999999999999999 >> .5?

I heard that you could right-shift a number by .5 instead of using Math.floor(). I decided to check its limits to make sure that it was a suitable replacement, so I checked the following values and got the following results in Google Chrome:
2.5 >> .5 == 2;
2.9999 >> .5 == 2;
2.999999999999999 >> .5 == 2; // 15 9s
2.9999999999999999 >> .5 == 3; // 16 9s
After some fiddling, I found out that the highest possible value of two which, when right-shifted by .5, would yield 2 is 2.9999999999999997779553950749686919152736663818359374999999¯ (with the 9 repeating) in Chrome and Firefox. The number is 2.9999999999999997779¯ in IE.
My question is: what is the significance of the number .0000000000000007779553950749686919152736663818359374? It's a very strange number and it really piqued my curiosity.
I've been trying to find an answer or at least some kind of pattern, but I think my problem lies in the fact that I really don't understand the bitwise operation. I understand the idea in principle, but shifting a bit sequence by .5 doesn't make any sense at all to me. Any help is appreciated.
For the record, the weird digit sequence changes with 2^x. The highest possible values of the following numbers that still truncate properly:
for 0: 0.9999999999999999444888487687421729788184165954589843749¯
for 1: 1.9999999999999999888977697537484345957636833190917968749¯
for 2-3: x+.99999999999999977795539507496869191527366638183593749¯
for 4-7: x+.9999999999999995559107901499373838305473327636718749¯
for 8-15: x+.999999999999999111821580299874767661094665527343749¯
...and so forth

Actually, you're simply ending up doing a floor() on the first operand, without any floating point operations going on. Since the left shift and right shift bitwise operations only make sense with integer operands, the JavaScript engine is converting the two operands to integers first:
2.999999 >> 0.5
Becomes:
Math.floor(2.999999) >> Math.floor(0.5)
Which in turn is:
2 >> 0
Shifting by 0 bits means "don't do a shift" and therefore you end up with the first operand, simply truncated to an integer.
The SpiderMonkey source code has:
switch (op) {
case JSOP_LSH:
case JSOP_RSH:
if (!js_DoubleToECMAInt32(cx, d, &i)) // Same as Math.floor()
return JS_FALSE;
if (!js_DoubleToECMAInt32(cx, d2, &j)) // Same as Math.floor()
return JS_FALSE;
j &= 31;
d = (op == JSOP_LSH) ? i << j : i >> j;
break;
Your seeing a "rounding up" with certain numbers is due to the fact the JavaScript engine can't handle decimal digits beyond a certain precision and therefore your number ends up getting rounded up to the next integer. Try this in your browser:
alert(2.999999999999999);
You'll get 2.999999999999999. Now try adding one more 9:
alert(2.9999999999999999);
You'll get a 3.

This is possibly the single worst idea I have ever seen. Its only possible purpose for existing is for winning an obfusticated code contest. There's no significance to the long numbers you posted -- they're an artifact of the underlying floating-point implementation, filtered through god-knows how many intermediate layers. Bit-shifting by a fractional number of bytes is insane and I'm surprised it doesn't raise an exception -- but that's Javascript, always willing to redefine "insane".
If I were you, I'd avoid ever using this "feature". Its only value is as a possible root cause for an unusual error condition. Use Math.floor() and take pity on the next programmer who will maintain the code.
Confirming a couple suspicions I had when reading the question:
Right-shifting any fractional number x by any fractional number y will simply truncate x, giving the same result as Math.floor() while thoroughly confusing the reader.
2.999999999999999777955395074968691915... is simply the largest number that can be differentiated from "3". Try evaluating it by itself -- if you add anything to it, it will evaluate to 3. This is an artifact of the browser and local system's floating-point implementation.

If you wanna go deeper, read "What Every Computer Scientist Should Know About Floating-Point Arithmetic": https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

Try this javascript out:
alert(parseFloat("2.9999999999999997779553950749686919152736663818359374999999"));
Then try this:
alert(parseFloat("2.9999999999999997779553950749686919152736663818359375"));
What you are seeing is simple floating point inaccuracy. For more information about that, see this for example: http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems.
The basic issue is that the closest that a floating point value can get to representing the second number is greater than or equal to 3, whereas the closes that the a float can get to the first number is strictly less than three.
As for why right shifting by 0.5 does anything sane at all, it seems that 0.5 is just itself getting converted to an int (0) beforehand. Then the original float (2.999...) is getting converted to an int by truncation, as usual.

I don't think your right shift is relevant. You are simply beyond the resolution of a double precision floating point constant.
In Chrome:
var x = 2.999999999999999777955395074968691915273666381835937499999;
var y = 2.9999999999999997779553950749686919152736663818359375;
document.write("x=" + x);
document.write(" y=" + y);
Prints out: x = 2.9999999999999996 y=3

The shift right operator only operates on integers (both sides). So, shifting right by .5 bits should be exactly equivalent to shifting right by 0 bits. And, the left hand side is converted to an integer before the shift operation, which does the same thing as Math.floor().

I suspect that converting 2.9999999999999997779553950749686919152736663818359374999999
to it's binary representation would be enlightening. It's probably only 1 bit different
from true 3.
Good guess, but no cigar.
As the double precision FP number has 53 bits, the last FP number before 3 is actually
(exact): 2.999999999999999555910790149937383830547332763671875
But why it is
2.9999999999999997779553950749686919152736663818359375
(and this is exact, not 49999... !)
which is higher than the last displayable unit ? Rounding. The conversion routine (String to number) simply is correctly programmed to round the input the the next floating point number.
2.999999999999999555910790149937383830547332763671875
.......(values between, increasing) -> round down
2.9999999999999997779553950749686919152736663818359375
....... (values between, increasing) -> round up to 3
3
The conversion input must use full precision. If the number is exactly the half between
those two fp numbers (which is 2.9999999999999997779553950749686919152736663818359375)
the rounding depends on the setted flags. The default rounding is round to even, meaning that the number will be rounded to the next even number.
Now
3 = 11. (binary)
2.999... = 10.11111111111...... (binary)
All bits are set, the number is always odd. That means that the exact half number will be rounded up, so you are getting the strange .....49999 period because it must be smaller than the exact half to be distinguishable from 3.

I suspect that converting 2.9999999999999997779553950749686919152736663818359374999999 to its binary representation would be enlightening. It's probably only 1 bit different from true 3.

And to add to John's answer, the odds of this being more performant than Math.floor are vanishingly small.
I don't know if JavaScript uses floating-point numbers or some kind of infinite-precision library, but either way, you're going to get rounding errors on an operation like this -- even if it's pretty well defined.

It should be noted that the number ".0000000000000007779553950749686919152736663818359374" is quite possibly the Epsilon, defined as "the smallest number E such that (1+E) > 1."