Why is foldl defined in a strange way in Racket?

Question

In Haskell, like in many other functional languages, the function foldl is defined such that, for example, foldl (-) 0 [1,2,3,4] = -10.

Thi

Answer 1

"differently than in any other language"

As a counter-example, Standard ML (ML is a very old and influential functional language)'s foldl also works this way: http://www.standardml.org/Basis/list.html#SIG:LIST.foldl:VAL