accumulate variant

[caduel]

The code is basically correct (assuming that's just part of it).
(If it is supposed to be complete: Other is all private, the constructors are missing etc.)


When I add that missing stuff (hint: do not forget to initialize _missed) the variable mean is assigned a meaningful value (namely the sum of my Values).

What is the problem you're having?

[mickey]

I need sum all "valid" values (exactly what it do) BUT I need to compute the "MEAN"; so I need to know how many "values" I've summed; how know how many they?

[caduel]

ok... you can do that by either:
* counting those not _missed (and thus iterating twice over your vector)
* or, by using a function object that counts both values and missed
(* or you could use globals, which is very bad imho)

try to replace your function by

class MeanAccumulator
{
	int counted_;
	double sum_;
public:
	MeanAccumulator() : counted_(0), sum_(0) {}
	void operator() (const Value &right)
        {
            if ( right.getMissed() ) return;
            ++counted_;
            sum_ += right._value;
         }
		double mean() const {return counted_ ? sum_/counted_ : 0; }
};
 
void computeMean() {
    MeanAccumulator res = std::for_each( _values.begin(), _values.end(),
                                         MeanAccumulator());
    std::cerr << "sum="<<res.mean()<<std::endl;
}

class MeanAccumulator
{
	int counted_;
	double sum_;
public:
	MeanAccumulator() : counted_(0), sum_(0) {}
	void operator() (const Value &right)
        {
            if ( right.getMissed() ) return;
            ++counted_;
            sum_ += right._value;
         }
		double mean() const {return counted_ ? sum_/counted_ : 0; }
};

void computeMean() {
    MeanAccumulator res = std::for_each( _values.begin(), _values.end(),
                                         MeanAccumulator());
    std::cerr << "sum="<<res.mean()<<std::endl;
}

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Note that the function object passed into for_each does NOT get modified.
The resulting function object (with the updated state) is returned by for_each.

(I assumed you do not insist on using accumulate.)

HTH