Fix behavior of foldr on empty arrays (#20144)

base/reduce.jl

@@ -131,7 +131,13 @@ mapfoldr(f, op, v0, itr) = mapfoldr_impl(f, op, v0, itr, endof(itr))
 Like `mapfoldr(f, op, v0, itr)`, but using the first element of `itr` as `v0`. In general,
 this cannot be used with empty collections (see `reduce(op, itr)`).
 """
-mapfoldr(f, op, itr) = (i = endof(itr); mapfoldr_impl(f, op, f(itr[i]), itr, i-1))
+function mapfoldr(f, op, itr)
+    i = endof(itr);
+    if i == 0
+        return Base.mr_empty_iter(f, op, itr, iteratoreltype(itr))
+    end
+    return mapfoldr_impl(f, op, f(itr[i]), itr, i-1)
+end
 
 """
     foldr(op, v0, itr)

test/reduce.jl

@@ -1,6 +1,7 @@
 # This file is a part of Julia. License is MIT: http://julialang.org/license
 
 # fold(l|r) & mapfold(l|r)
+@test foldl(+, Int64[]) == 0
 @test foldl(-, 1:5) == -13
 @test foldl(-, 10, 1:5) == -5
 
@@ -16,6 +17,7 @@
 @test Base.mapfoldl((x)-> x ⊻ true, |, [true false true false false]) == true
 @test Base.mapfoldl((x)-> x ⊻ true, |, false, [true false true false false]) == true
 
+@test foldr(+, Int64[]) == 0
 @test foldr(-, 1:5) == 3
 @test foldr(-, 10, 1:5) == -7