Integer arithmetic

[Daniel Timothy Bentley]

Am I so far off in thinking maybe this should be provided by the language?
Some way to create an int, using some (perhaps implementation-defined)
method of changing a long to an int?  Some way to ensure that I don't
misguess on what themask is?  What if I'm programming on a machine with 36
bit longs?  Or where 64 bits is the atomic datatype?
It seems like there should be an int constructor that won't overflow in
this case.

Unless there's a function I'm missing that returns INT_MAX (as C
pronounces it) and I'm just supposed to mod it.

-Dan

On Tue, 25 Mar 2003, Alex Martelli wrote:

> Daniel Timothy Bentley wrote:
>
> > I am currently embarking on the utterly futile and mildly recursive task
> > of implementing a C interpreter in python (Jython specifically).
> >
> > Right now, I'm trying to think about integer ops.
> >
> > I want to add two integers, e.g., and have the result be an integer.  The
>
> And what do you want to happen when the two integers you're adding yield
> a result too large to be represented as an integer?  You do no specify
> what behavior you want for that (and I believe the C standard doesn't,
> either, though I'm not familiar with the 1999 version, only the older one).
>
> Basically what you need is a function makeint that will return the int
> of its single argument -- if that overflows, then you may want to e.g.
> truncate that to 32 bits, or whatever.  Since most cases won't overflow
> you're best off with an "easier to ask forgiveness than permission"
> structure, e.g.:
>
> def mask(x, themask = 0x7FFFFFFF):
>     return int(x & themask)
>
> def makeint(x):
>     try: return int(x)
>     except OverflowError:
>         if x < 0:
>             return -mask(-x)
>         else:
>             return mask(x)
>
> > naive way:
> >
> > foo = bar + baz
> >
> > Doesn't work because of automatic promotion to bignum.
>
> so you code
>
>     foo = makeint(bar + baz)
>
> with some makeint function such as the above one.
>
>
> Alex
>
>