foldr(op, itr)

..  foldr(op, itr)

Like ``foldr(op, v0, itr)``, but using the last element of ``itr``
as ``v0``. In general, this cannot be used with empty collections
(see ``reduce(op, itr)``).


In the Julia programming language, the function foldr(op, v0, itr) is similar to the reduce function, but with guaranteed right associativity. It takes an operator op, an initial value v0, and an iterable itr as arguments. The operator is applied to each element of the iterable from right to left, using the initial value as the starting point.

julia> foldr(+, 0, [1, 2, 3, 4])

julia> foldr(string, "", ["Hello", " ", "World", "!"])
"!World Hello"
  1. Summing elements in an array:

    julia> arr = [1, 2, 3, 4, 5];
    julia> foldr(+, 0, arr)

    This example calculates the sum of all elements in the array arr using the + operator and an initial value of 0.

  2. Concatenating strings:

    julia> strings = ["Hello", " ", "Julia"];
    julia> foldr(string, "", strings)
    "Julia Hello "

    It concatenates the strings in the strings array from right to left, using an initial value of an empty string.

  3. Applying a custom function:
    julia> double(x, y) = 2x + y;
    julia> foldr(double, 0, [1, 2, 3, 4])

    In this example, the custom function double is applied to each element of the array, accumulating the result from right to left.

Common mistake example:

julia> foldr(-, 0, [1, 2, 3, 4])
ERROR: MethodError: no method matching foldr(::typeof(-), ::Int64, ::Array{Int64,1})

In this case, the - operator cannot be directly used with foldr because it requires two arguments. To fix this, you can define a custom function or use an anonymous function to perform the subtraction:

julia> foldr((x, y) -> x - y, 0, [1, 2, 3, 4])

Ensure that the operator or function you provide to foldr can handle the right-to-left accumulation correctly.

See Also

foldl, foldr, mapfoldl, mapfoldr, mapreduce, mapreducedim,

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