Homework: Cout incorrectly handling a return value of 0.0 from a class method

First, this is homework, so I can not dynamically allocate memory for an array of any size, and I can not use a vector.

I have a class which includes a double array holding 30 elements, and two other variables to keep track of how many elements have been added and the max number of elements that can be st开发者_C百科ored.

There are several methods that return the highest, lowest, average, and total from the elements in the array. An example of one of the methods is...

double Stats::sum() const
{
  double sum = 0.0;

  for (unsigned short i = 0; i < nElements; ++i)
    sum += stats[i];

  return sum;
};

In my main() function I have a cout statement...

cout << "\nTotal rainfall for " << MonthlyRainfall.size() << " months is "
     << MonthlyRainfall.sum() << " inches.\n";

When there are values in the array, the output is what I expect...

Total rainfall for 1 months is 1.5 inches.

However, when there are no values in the array (the method returns 0.0), but the output looks like...

Total rainfall for 0 months is -1.$ inches.

Can anyone help me understand what's happening in the cout statement that's causing the 0.0 returned by my method to output as it is?

Note: At the beginning of the main() function, the following statement is executed to format the decimal output. cout << fixed << showpoint << setprecision(1);

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Update I guess it was late and I was calling the average() method instead of sum(). I have fixed it and you also pointed out that I needed to make a couple changes to the average method to ensure a divide by 0 is not happening. (=

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Divide-by-zero is well-defined to be positive or negative infinity. By default this displays as

Inf

or

-Inf

(at least on my machine).

setprecison messes this up, see this other question.

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average() has divided something by zero and got minus infinity, and that's how cout has displayed it.

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You failed to show the average method but my guess is that you do a divide by zero in there. Wackiness ensues.