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Re: "i = i|0"

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Am 12.06.2014 00:20, schrieb Thomas 'PointedEars' Lahn:

> First of all, this does _not_ express the C semantics in JavaScript, and
> there is no other way.  “int” is a *generic* type in C/C++; IIUC, the result
> could be a 32-bit integer when compiled for a 32-bit platform or a 64-bit
> integer when compiled for a 64-bit platform.

Not quite. int is a type whatever the compiler chooses it to be, let it 
be 32 bit, 64 bit or 36 bit, no matter what the platform is. I have seen 
platforms were int was a 16 bit type even tough the platform was 32 bit 
wide.

IOWs, the C to JavaScript compiler can surely implement a valid C model 
on top of JavaScript with 32 bit ints, the decision to have ints 32 bit 
wide is not invalidated by the platform having any other native register 
size.

> By contrast, using the binary bitwise OR operator, as with all ECMAScript
> binary bitwise operators, *always* creates an IEEE-754 double-precision
> *floating-point* value representing a *32-bit* integer value.  Also, not
> only the result will be such a value, but also the operands are converted
> to such values internally, before the operation is performed.

Within the limits of what JavaScript has to offer, this is the closest 
it can do to emulate the C type system when operating on signed 
datatypes; these are also first converted to int (here emulated as 32 
bit types), then the operation is performed. It can get very tricky to 
emulate the *unsigned* types, and I wonder what the C to JavaScript 
compiler would do about them.


> Conversion to an integer value of the ECMAScript Number type (i.e., where
> the fractional part of the mantissa is zero), which appears to be the goal
> here, can be better achieved with the Math.floor() and Math.ceil()
> functions, e.g.:

No, I don't think that this is the goal here, i.e. rounding to an 
integer number (avoiding the term "int" here intentionally). An "int" 
has undefined behavior if an operation overflows. Just rounding would 
create another integer number, but one that is outside of the bit range, 
possibly with even weirder side effects in the program. That's possibly 
not quite what was intended. In the end, it probably does not matter 
much because, as said, C leaves it undefined what happens on overflow of 
the signed types, and the implementation just picked the simplest 
possible choice to create a "round to zero".

> Of course, this still does not remotely implement the C semantics.  One
> aspect of it is that code where you pass a non-integer would not compile.

Not so. Instead, C would truncate the non-integer to the next integer 
(cut off the fractional part) and would use that as input. The following 
compiles fine:

int plusone(int x)
{
   return x+1;
}

int main(int argc,char **argv)
{

   return plusone(2.2);
}

I'm not sure how JavaScript handles negative values when applied to the 
binary or operator, but I would suspect that it also performs a "round 
to zero", which is exactly what C does, thus a perfect match. That is 
possibly even intentional in JavaScript.

> Since ECMAScript uses dynamic type-checking, it is not possible to prevent
> compilation.  But at the very least passing unsuitable values should cause
> an exception to be thrown, so that it becomes unnecessary to handle them,
> e.g.:

Certainly not, since C does not have exceptions either. You could do 
that possibly on overflow (with an impact on runtime performance), but 
not on passing in a non-integer though numeric type.

Greetings,

Thomas

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Re: "i = i|0" Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2014-06-12 00:20 +0200
  Re: "i = i|0" raltbos@xs4all.nl (Richard Bos) - 2014-06-12 11:33 +0000
    Re: "i = i|0" Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2014-06-12 14:28 +0200
  Re: "i = i|0" Thomas Richter <thor@math.tu-berlin.de> - 2014-06-12 19:52 +0200
    Re: "i = i|0" Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2014-06-12 20:25 +0200

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