The answer
Yes. Each sum(a) call creates a new function with a new closure over a. The previous closure is replaced, but the accumulated value (a + b) is passed forward to the new sum call.
How the sum(1)(2)(3) Function Uses Closures
Does each call to sum create a new closure?
More FAQs in How the sum(1)(2)(3) Function Uses Closures
Each call to sum(a) returns a function that closes over a. When called with b, it returns sum(a + b), creating a new closure with the accumulated total. The closure keeps the running sum alive between calls.
b is undefined. The function checks if (b !== undefined), which is false. It returns a (the accumulated sum). This is how the chain terminates and the result is retrieved.
No. The curried sum requires each returned function to remember the accumulated total. This is only possible with closures. Without closures, the inner function would not have access to the previous arguments.
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