foldl'
lib.lists.foldl' (Prime)
Reduce a list by applying a binary operator from left to right, starting with an initial accumulator.
Before each application of the operator, the accumulator value is evaluated.
This behavior makes this function stricter than foldl
.
Unlike builtins.foldl'
,
the initial accumulator argument is evaluated before the first iteration.
A call like
foldl' op acc₀ [ x₀ x₁ x₂ ... xₙ₋₁ xₙ ]
is (denotationally) equivalent to the following,
but with the added benefit that foldl'
itself will never overflow the stack.
let
acc₁ = builtins.seq acc₀ (op acc₀ x₀ );
acc₂ = builtins.seq acc₁ (op acc₁ x₁ );
acc₃ = builtins.seq acc₂ (op acc₂ x₂ );
...
accₙ = builtins.seq accₙ₋₁ (op accₙ₋₁ xₙ₋₁);
accₙ₊₁ = builtins.seq accₙ (op accₙ xₙ );
in
accₙ₊₁
# Or ignoring builtins.seq
op (op (... (op (op (op acc₀ x₀) x₁) x₂) ...) xₙ₋₁) xₙ
Inputs
op

The binary operation to run, where the two arguments are:
acc
: The current accumulator value: Either the initial one for the first iteration, or the result of the previous iterationx
: The corresponding list element for this iteration
acc

The initial accumulator value.
The accumulator value is evaluated in any case before the first iteration starts.
To avoid evaluation even before the
list
argument is given an eta expansion can be used:list: lib.foldl' op acc list
list

The list to fold
Type
foldl' :: (acc > x > acc) > acc > [x] > acc
Examples
lib.lists.foldl'
usage example
foldl' (acc: x: acc + x) 0 [1 2 3]
=> 6