Understanding std::accumulate

You're making a mistaken assumption: that type T is of the same type as the InputIterator.

But std::accumulate is generic, and allows all different kinds of creative accumulations and reductions.

Example #1: Accumulate salary across Employees

Here's a simple example: an Employee class, with many data fields.

class Employee {
/** All kinds of data: name, ID number, phone, email address... */
public:
 int monthlyPay() const;
};

You can't meaningfully "accumulate" a set of employees. That makes no sense; it's undefined. But, you can define an accumulation regarding the employees. Let's say we want to sum up all the monthly pay of all employees. std::accumulate can do that:

/** Simple class defining how to add a single Employee's
 *  monthly pay to our existing tally */
auto accumulate_func = [](int accumulator, const Employee& emp) {
   return accumulator + emp.monthlyPay();
 };

// And here's how you call the actual calculation:
int TotalMonthlyPayrollCost(const vector<Employee>& V)
{
 return std::accumulate(V.begin(), V.end(), 0, accumulate_func);
}

So in this example, we're accumulating an int value over a collection of Employee objects. Here, the accumulation sum isn't the same type of variable that we're actually summing over.

Example #2: Accumulating an average

You can use accumulate for more complex types of accumulations as well - maybe want to append values to a vector; maybe you have some arcane statistic you're tracking across the input; etc. What you accumulate doesn't have to be just a number; it can be something more complex.

For example, here's a simple example of using accumulate to calculate the average of a vector of ints:

// This time our accumulator isn't an int -- it's a structure that lets us
// accumulate an average.
struct average_accumulate_t
{
    int sum;
    size_t n;
    double GetAverage() const { return ((double)sum)/n; }
};

// Here's HOW we add a value to the average:
auto func_accumulate_average = 
    [](average_accumulate_t accAverage, int value) {
        return average_accumulate_t(
            {accAverage.sum+value, // value is added to the total sum
            accAverage.n+1});      // increment number of values seen
    };

double CalculateAverage(const vector<int>& V)
{
    average_accumulate_t res =
        std::accumulate(V.begin(), V.end(), average_accumulate_t({0,0}), func_accumulate_average)
    return res.GetAverage();
}

Example #3: Accumulate a running average

Another reason you need the initial value is because that value isn't always the default/neutral value for the calculation you're making.

Let's build on the average example we've already seen. But now, we want a class that can hold a running average -- that is, we can keep feeding in new values, and check the average so far, across multiple calls.

class RunningAverage
{
    average_accumulate_t _avg;
public:
    RunningAverage():_avg({0,0}){} // initialize to empty average

    double AverageSoFar() const { return _avg.GetAverage(); }

    void AddValues(const vector<int>& v)
    {
        _avg = std::accumulate(v.begin(), v.end(), 
            _avg, // NOT the default initial {0,0}!
            func_accumulate_average);
    }

};

int main()
{
    RunningAverage r;
    r.AddValues(vector<int>({1,1,1}));
    std::cout << "Running Average: " << r.AverageSoFar() << std::endl; // 1.0
    r.AddValues(vector<int>({-1,-1,-1}));
    std::cout << "Running Average: " << r.AverageSoFar() << std::endl; // 0.0
}

This is a case where we absolutely rely on being able to set that initial value for std::accumulate - we need to be able to initialize the accumulation from different starting points.

---

In summary, std::accumulate is good for any time you're iterating over an input range, and building up one single result across that range. But the result doesn't need to be the same type as the range, and you can't make any assumptions about what initial value to use -- which is why you must have an initial instance to use as the accumulating result.

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.

If you wanted accumulate(V.begin()+1, V.end(), V.begin()) you could just write that. But what if you thought v.begin() might be v.end() (i.e. v is empty)? What if v.begin() + 1 is not implemented (because v only implements ++, not generized addition)? What if the type of the accumulator is not the type of the elements? Eg.

std::accumulate(v.begin(), v.end(), 0, [](long count, char c){
   return isalpha(c) ? count + 1 : count
});