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Fixed dace::math::pow
for Integer case
#1748
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philip-paul-mueller
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8f4e7df
Added a pow specialization for integer.
philip-paul-mueller eb887d9
Enabled the FFT test again.
philip-paul-mueller 4357ee5
Merge branch 'main' into better_pow
tbennun e7d7990
Fixed some errors in the power implementation.
philip-paul-mueller db4b411
Some better types.
philip-paul-mueller 238f15d
Fixed a bug in `ipow`.
philip-paul-mueller d355c12
Made `ipow` also more general.
philip-paul-mueller def4d42
Added a specialization of `ipow` for signed integers.
philip-paul-mueller 28feda5
Again fixed `ipow` now it was a redeclaration.
philip-paul-mueller 28efc8c
Let's try this.
philip-paul-mueller 9773ee8
Unfixed `ipow`.
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So, if both a and b are integers (signed or unsigned), this path will be followed? In other words, if we want to compute, e.g., 3^-5, we need to convert one of them to float. Are we sure this is the behavior we want?
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This is correct.
But we kind of need this behaviour, to enable the FFT test this commit enables.
There the problem is that a size of an array is given as
a = np.ones(2**3)
.Which the code generator translates to
If
pow
would return a double, then it is a compilation error.There are technically three things we can do:
new[]
expressions are always casted tostd::size_t
.However, this would be another PR.
What is your opinion on this @tbennun ?
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I agree with @alexnick83. The reason we didn't have these pow implementations in the first place was because of negative exponents. I think we should use
ipow
when we know something or the equivalent version only in the case of an unsigned exponent and integer base.There was a problem hiding this comment.
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I looked it up in the code generator,
ipow
is only used if the exponent is known at code generation time.This is not the case in the FFT test, there the exponent is not known and thus
pow
is selected.The reason why it lead to the compilation error in the FFT test was because the arguments of
pow
whereint64
and since there is only an overload forint
andunsigned
int, the template was selected.However, even before the expression
5^-3
would have resulted in the value 0, given that both were integer.For the reasons outlined above, I do not think that this PR changes anything big.
But how should I proceed @alexnick83 @tbennun
PS: I also tried to implement
ipow
for negative exponents and that failed with some compilation bug related tohalf
, which I did not understood.