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How to declare a local array #5
Description
The requirement is that it works with CPython and NumPy, but also compiles using LPython. One option is:
def main():
a: f32[10000] = numpy.empty(10000, dtype=numpy.float32)
b: f32[10000] = numpy.empty(10000, dtype=numpy.float32)
c: f32[10000] = numpy.empty(10000, dtype=numpy.float32)
scalar: f32
i: i32
nsize: i32
scalar = 10
nsize = size(a)
for i in range(nsize): # type: parallel
a[i] = 5
b[i] = 5
triad(a, b, scalar, c)
print("End Stream Triad")
So the syntax is:
a: f32[10000] = numpy.empty(10000, dtype=numpy.float32)
But this is redundant, as the size 10000 is repeated twice and the type f32 and numpy.float32 is also repeated.
Activity
certik commentedon Jan 5, 2022
One way that would definitely work is to make LPython infer the type and array size from the RHS from a code like this:
The alternative, that is, to declare arrays as:
and somehow make it work with CPython is unclear, as somehow the type annotation would need to "execute" and call NumPy.
certik commentedon Jan 6, 2022
After discussing with @namannimmo10, I think the best way is to use:
or an equivalent:
Both of which work in CPython/NumPy. This follows LPython's initial design that requires everything to be explicitly typed. In both cases,
ais "declared" as a local array and allocated (on stack or heap, that is ASR backend dependent).Later on, as we implement type inference, we can omit the type annotation, and just do:
and the compiler will infer the type
f32[10000]forafrom the RHS.If the array is only typed using
a: f32[10000]but never assigned to usinga = numpy.empty(10000, dtype=numpy.float32)or similar, then LPython should given an error message.certik commentedon Jan 7, 2022
Here is another example. This should give a compiler error, that
bis not allocated:The way to write it is as:
or like this:
Another example: this should work, and
bis a pointer toa.certik commentedon Mar 21, 2023
This is mostly fixed, I opened up an issue with the remaining bug: #1599