Develop Reference

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.

Working with string (array?) of bits of an unspecified length

I'm a javascript code monkey, so this is virgin territory for me.
I have two "strings" that are just zeros and ones:
var first = "00110101011101010010101110100101010101010101010";
var second = "11001010100010101101010001011010101010101010101";
I want to perform a bitwise & (which I've never before worked with) to determine if there's any index where 1 appears in both strings.
These could potentially be VERY long strings (in the thousands of characters). I thought about adding them together as numbers, then converting to strings and checking for a 2, but javascript can't hold precision in large intervals and I get back numbers as strings like "1.1111111118215729e+95", which doesn't really do me much good.
Can I take two strings of unspecified length (they may not be the same length either) and somehow use a bitwise & to compare them?
I've already built the loop-through-each-character solution, but 1001^0110 would strike me as a major performance upgrade. Please do not give the javascript looping solution as an answer, this question is about using bitwise operators.

As you already noticed yourself, javascript has limited capabilities if it's about integer values. You'll have to chop your strings into "edible" portions and work your way through them. Since the parseInt() function accepts a base, you could convert 64 characters to an 8 byte int (or 32 to a 4 byte int) and use an and-operator to test for set bits (if (a & b != 0))

var first = "00110101011101010010101110100101010101010101010010001001010001010100011111",
second = "10110101011101010010101110100101010101010101010010001001010001010100011100",
firstInt = parseInt(first, 2),
secondInt = parseInt(second, 2),
xorResult = firstInt ^ secondInt, //524288
xorString = xorResult.toString(2); //"10000000000000000000"

why is 10 * (7/10) = 7 in JavaScript?

7 and 10 in the expression (7/10) are integers, so the result 0.7 should be integer as well, which is 0, and the result for the entire expression should be 0 too. However, it's giving me the result of 7, why? Is it ignoring the parentheses or converts to double automatically?

JavaScript doesn't distinguish between integers and floating point numbers, everything I believe is considered a double so that is just why you get the result.
Take a look at the details on the Number property on MDN.

JavaScript doesn't have an integer type, or a double, or a float... it just has 1 type for all numbers: the helpfuly called Number type (try var foo = new Number(7);, or var foo = Number('123string');
Now, I know I said that JS doesn't know of floats, but that's not entirely true. All Number type vars/values are, essentially 64 bit floats, as defined by the IEEE 754 standard (which are, indeed, as Jan Dvorak kindly pointed out to me, double's in most staticly typed languages), with all the caveats that brings with it:
(.1 + .2);//0.30000000000000004
But that's not the point. The point is that, in JS you can perform float + int arithmatic without there ever being a need for internal casts, or conversions. That's why 10*(7/10) will always be 7

There is no int and double in JavaScript
In JavaScript, both int, flot, and double are normalized to work together. They are treated as 1 (They're treated as as Number, which is an IEEE 754 float. Thanks #Elias Van Ootegem). Equality, Liberty and Fraternity. and thus;
10*0.7 = 7
JavaScript is not like C.

Javascript doesn't have integers, and even if it did, there's nothing that says that / needs to return an integer (just because another language may do that doesn't mean every language has to). The operation results in a float/Number, just like all Javascript numbers are, period.

try this
10*parseInt(7/10)
hope this will help you

If you try to follow the rules, then
10 * (7/10) --> 10 * .7 --> 7
You cannot change the way its gonna result into.
so the result 0.7 should be integer as well, which is 0
If you want this, then try using
Math.Floor();
This would change the decimals to the nearest int! Or try out parse()

JavaScript uses dynamic types. That means that a variable like this:
var str = "hi";
Can later become:
str = 123; //now we have an 'int'
str += 0.35; //now str is 123.35, a 'float'
So JavaScript doesn't cast floats to ints for example.
If you want to force a "cast" then you have to do:
var integer = parseInt( 3.14*9.0291+23, 10 ); //the second parameter (10) is the 'base'
But remember, Javascript will not take care of types, that's your problem.

Parse json in javascript - long numbers get rounded

I need to parse a json that contains a long number (that was produces in a java servlet). The problem is the long number gets rounded.
When this code is executed:
var s = '{"x":6855337641038665531}';
var obj = JSON.parse(s);
alert (obj.x);
the output is:
6855337641038666000
see an example here: http://jsfiddle.net/huqUh/
why is that, and how can I solve it?

As others have stated, this is because the number is too big. However, you can work around this limitation by sending the number as a string like so:
var s = '{"x":"6855337641038665531"}';
Then instead of using JSON.parse(), you can use a library such as javascript-bignum to work with the number.

It's too big of a number. JavaScript uses double-precision floats for numbers, and they have about 15 digits of precision (in base 10). The highest integer that JavaScript can reliably save is something like 251.
The solution is to use reasonable numbers. There is no real way to handle such large numbers.

The largest number JavaScript can handle without loss of precision is 9007199254740992.

I faced this issue some time ago, I was able to solve using this lib: https://github.com/josdejong/lossless-json
You can check this example:
let text = '{"normal":2.3,"long":123456789012345678901,"big":2.3e+500}';
// JSON.parse will lose some digits and a whole number:
console.log(JSON.stringify(JSON.parse(text)));
// '{"normal":2.3,"long":123456789012345680000,"big":null}' WHOOPS!!!
// LosslessJSON.parse will preserve big numbers:
console.log(LosslessJSON.stringify(LosslessJSON.parse(text)));
// '{"normal":2.3,"long":123456789012345678901,"big":2.3e+500}'

Bitwise check in Javascript

I play a game, and in the database we set 100663296 to be a GM Leader but also this field in the database gets written to for different things, so it changes that number to 100794368
i was told to possible use a bit-wise check to check whether the first number is the same as the second number, and I have googled on using bit-wise checks but got confused as to what to use for my check.
Here are some other numbers that change, including the one from above.
predefined number new changed number/ever changing number.
100663296 = 100794368
67108864 = 67239936
117440512 = 2231767040
so how should i go about checking these numbers?
And here is part of my code that i was using before i noticed the change in the numbers.
if (playerData[i].nameflags == 67108864)
{
playerRows += '<img src ="icons/GM-Icon.png" alt="GM" title="GM"></img>';
}
thx to Bergi, for the answer.
if (playerData[i].nameflags & 0x400000 /* === 0x400000 */)
this seams to work great.
also thx to vr1911428
and every one else for the help on this.

So let's convert those numbers to binary representation (unsigned integer):
> 100663296
00000110000000000000000000000000
> 100794368
00000110000000100000000000000000
> 67108864
00000100000000000000000000000000
> 67239936
00000100000000100000000000000000
> 117440512
00000111000000000000000000000000
> 2231767040
10000101000001100001000000000000
Notice that the last number is out of the scope of JavaScripts bitwise arithmetic, which only works with 32-bit signed integers - you won't be able to use the leftmost bit.
So which bits do you want to compare now? There are lots of possibilities, the above scheme doesn't make it clear, yet it looks like you are looking for the 27th bit from the right (226 = 67108864). To match against it, you can apply a binary AND bitmask:
x & Math.pow(2, 26)
which should evaluate to 226 again or zero - so you can just check for truthiness. Btw, instead of using pow you could use hexadecimal notation: 0x4000000. With that, your condition will look like this:
if (playerData[i].nameflags & 0x400000 /* === 0x400000 */)

If you need to check for full bitwise equality of two integers, all you need is just '==' operator, but to use it, you should guarantee that both operands are integers:
left = 12323;
right = 12323;
if (left == right)
alert("Operands are binary equal; I'll guarantee that. :-)");
Be very careful though; if at least one of operands is string representing number, not a number, both operands will be considered strings and you can get confusing results:
left = "012323";
right = 12323;
if (left != right)
alert("Operands are not equal, even though they represent 'equal values', as they are compared as strings.");
In general, these days, the attempt to operate with strings representing data instead of data itself is a real curse of the beginners; and it's hard to explain to them. It is especially difficult to explain in JavaScript, with its loose-type typing concept, which is itself very complex and hard to understand, behind the illusory simplicity.
Finally, if you need to compare separate bits (and, from your question, I don't see this need), you can use binary operators:
https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Bitwise_Operators
That's it, basically.
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