Leonid Volnitsky - 1 year ago 138

C++ Question

I want to know why

`std::accumulate`

`accumulate`

`vector<int> V{1,2,3};`

int sum = accumulate(V.begin(), V.end(), 0);

// sum == 6

Call to

`accumulate`

`sum = 0; // 0 - value of 3rd param`

for (auto x : V) sum += x;

There is also optional 4th parameter, which allow to replace addition with any other operation.

Rationale that I've heard is that if you need let say not to add up, but multiply elements of a vector, we need other (non-zero) initial value:

`vector<int> V{1,2,3};`

int product = accumulate(V.begin(), V.end(), 1, multiplies<int>());

But why not do like Python - set initial value for

`V.begin()`

`V.begin()+1`

`int sum = accumulate(V.begin()+1, V.end(), V.begin());`

This will work for any op. Why is 3rd parameter needed at all?

Answer Source

The way things are, it is annoying for code that knows for sure a range isn't empty and that wants to start accumulating from the first element of the range on. Depending on the operation that is used to accumulate with, it's not always obvious what the 'zero' value to use is.

If on the other hand you only provide a version that requires non-empty ranges, it's annoying for callers that don't know for sure that their ranges aren't empty. An additional burden is put on them.

One perspective is that the best of both worlds is of course to provide both functionality. As an example, Haskell provides both `foldl1`

and `foldr1`

(which require non-empty lists) alongside `foldl`

and `foldr`

(which mirror `std::transform`

).

Another perspective is that since the one can be implemented in terms of the other with a trivial transformation (as you've demonstrated: `std::transform(std::next(b), e, *b, f)`

-- `std::next`

is C++11 but the point still stands), it is preferable to make the interface as minimal as it can be with no real loss of expressive power.