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primops: make nature of foldl' strictness clearer

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* Clarify the documentation of foldl': That the arguments are forced
  before application (?) of `op` is necessarily true. What is important
  to stress is that we force every application of `op`, even when the
  value turns out to be unused.

* Move the example before the comment about strictness to make it less
  confusing: It is a general example and doesn't really showcase anything
  about foldl' strictness.

* Add test cases which nail down aspects of foldl' strictness:
  * The initial accumulator value is not forced unconditionally.
  * Applications of op are forced.
  * The list elements are not forced unconditionally.
This commit is contained in:
sternenseemann 2022-10-16 01:32:55 +02:00
parent ac0fb38e8a
commit d0f2da214b
6 changed files with 25 additions and 3 deletions
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6
src/libexpr/primops.cc
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@ -2908,9 +2908,9 @@ static RegisterPrimOp primop_foldlStrict({
.doc = R"(
Reduce a list by applying a binary operator, from left to right,
e.g. `foldl' op nul [x0 x1 x2 ...] = op (op (op nul x0) x1) x2)
...`. The operator is applied strictly, i.e., its arguments are
evaluated first. For example, `foldl' (x: y: x + y) 0 [1 2 3]`
evaluates to 6.
...`. For example, `foldl' (x: y: x + y) 0 [1 2 3]` evaluates to 6.
The return value of each application of `op` is evaluated immediately,
even for intermediate values.
)",
.fun = prim_foldlStrict,
});