4 ms·

Oh, and a question – is the generated code actually semantically identical? For most uses practical uses, it is, but I think under racy conditions it could give

Oh, and a question – is the generated code actually semantically identical? For most uses practical uses, it is, but I think under racy conditions it could give different results.

ygra 10y ago
It is, for the example. But it only works because it is known at that point that the input is not just an IEnumerable<int>, but rather an int[]. Knowing that they can generate a for loop, which is quite a bit faster than the enumerator. For an unknown IEnumerable they can still emit the loop and static helper method and cut down on the generated closure, but they cannot remove the IEnumerator usage. They could probably do the same as for the array if it's known that the source IEnumerable is a List<T> (and not a subclass, because GetEnumerator may have been overridden with side-effects). Effectively this transformation is the same as doing "Convert LINQ to code" and "Extract method" in R# (although those tend to sometimes be unable to codify even simple LINQ expressions).
- nhebb 10y ago
  >It is, for the example. It doesn't even look right for the example. This line: num += num2 + 3; Should be: num += _linqitems[i] + 3;
  - OskarS 10y ago
    Yeah, what the hell? That wont even compile. Is this thing even real? Why wouldn't the just use the actual output of their tool?
    - matthewwarren 10y ago
      It's just a typo on the README and it's been fixed now. The tool works fine, I've tried it out myself
      - OskarS 10y ago
        Ok, good. A thing like that doesn't inspire confidence, but presumably it was just a typo then.
  - ygra 10y ago
    They probably manually inlined a line like int num2 = _linqitems[i]; from the decompiled output. I'm fairly certain that if the transformation wouldn't result in compiling code, the compiler would refuse to compile it, resulting in no assembly to decompile. They also noted somewhere to have run the System.Linq tests and only one edge case involving Min/Max and NaN failed. Tests likely wouldn't run either if the code didn't compile.

Oh, and a question – is the generated code actually semantically identical? For most uses practical uses, it is, but I think under racy conditions it could give different results.

ygra 10y ago

It is, for the example. But it only works because it is known at that point that the input is not just an IEnumerable<int>, but rather an int[]. Knowing that they can generate a for loop, which is quite a bit faster than the enumerator. For an unknown IEnumerable they can still emit the loop and static helper method and cut down on the generated closure, but they cannot remove the IEnumerator usage. They could probably do the same as for the array if it's known that the source IEnumerable is a List<T> (and not a subclass, because GetEnumerator may have been overridden with side-effects). Effectively this transformation is the same as doing "Convert LINQ to code" and "Extract method" in R# (although those tend to sometimes be unable to codify even simple LINQ expressions).

nhebb 10y ago
>It is, for the example. It doesn't even look right for the example. This line: num += num2 + 3; Should be: num += _linqitems[i] + 3;
- OskarS 10y ago
  Yeah, what the hell? That wont even compile. Is this thing even real? Why wouldn't the just use the actual output of their tool?
  - matthewwarren 10y ago
    It's just a typo on the README and it's been fixed now. The tool works fine, I've tried it out myself
    - OskarS 10y ago
      Ok, good. A thing like that doesn't inspire confidence, but presumably it was just a typo then.
- ygra 10y ago
  They probably manually inlined a line like int num2 = _linqitems[i]; from the decompiled output. I'm fairly certain that if the transformation wouldn't result in compiling code, the compiler would refuse to compile it, resulting in no assembly to decompile. They also noted somewhere to have run the System.Linq tests and only one edge case involving Min/Max and NaN failed. Tests likely wouldn't run either if the code didn't compile.