Computation
3.6.3 A numeric example
Let’s look at a more realistic example. Often, we have a series of values that we want to read into our program so that we can do something with them.
The “something” could be producing a graph of the values, calculating the mean and median, finding the largest element, sorting them, combining them with other data, searching for “interesting” values, comparing them to other data, etc. There is no limit to the range of computations we might perform on data, but first we need to get it into our computer’s memory. Here is the basic technique for getting an unknown – possibly large – amount of data into a computer. As a concrete example, we chose to read in floating-point numbers representing temperatures:
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int main()
// read some temperatures into a vector {
vector<double> temps; // temperatures for(double temp; cin>>temp;) // read into temp temps.push_back(temp); // put temp into vector // ...do something . . .
}
So, what goes on here? First we declare a vector to hold the data:
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vector<double> temps; // temperatures
This is where the type of input we expect is mentioned. We read and store
doubles.
Next comes the actual read loop:
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for (double temp; cin>>temp; ) // read into temp temps.push_back(temp); // put temp into vector
We define a variable temp of type double to read into. The cin>>temp reads a
double, and that double is pushed into the vector (placed at the back). We have seen those individual operations before. What’s new here is that we use the input operation, cin>>temp, as the condition for a for-statement. Basically,
cin>>temp is true if a value was read correctly and false otherwise, so that for- statement will read all the doubles we give it and stop when we give it
anything else. For example, if you typed
1.2 3.4 5.6 7.8 9.0 |
then temps would get the five elements 1.2, 3.4, 5.6, 7.8, 9.0 (in that order, for example, temps[0]==1.2). We used the character '|' to terminate the input – anything that isn’t a double can be used. In §9.4 we discuss how to terminate input and how to deal with errors in input.
To limit the scope of our input variable, temp, to the loop, we used a for- statement, rather than a while-statement:
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double temp;
while (cin>>temp) // read
temps.push_back(temp); // put into vector // ... temp might be used here ...
As usual, a for-loop shows what is going on “up front” so that the code is easier to understand and accidental errors are harder to make.
Once we get data into a vector we can easily manipulate it. As an example, let’s calculate the mean and median temperatures:
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int main()
// compute mean and median temperatures
{
vector<double> temps; // temperatures for(double temp; cin>>temp;) // read into temp temps.push_back(temp); // put temp into vector
// compute mean temperature:
double sum = 0;
for (double x : temps) sum += x;
cout << "Average temperature: " << sum/temps.size() << '\n';
// compute median temperature:
ranges::sort(temps); // sort the temperatures cout << "Median temperature: " << temps[temps.size()/2] << '\n';
}
We calculate the average (the mean) by simply adding all the elements into
sum, and then dividing the sum by the number of elements (that is,
temps.size()):
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// compute average temperature:
double sum = 0;
for (double x : temps) sum += x;
cout << "Average temperature: " << sum/temps.size() << '\n';
Note how the += operator comes in handy.
To calculate a median (a value chosen so that half of the values are lower and the other half are higher) we need to sort the elements. For that, we use a variant of the standard-library sort algorithm, sort():
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// compute median temperature:
ranges::sort(temps); // sort the temperatures cout << "Median temperature: " << temps[temps.size()/2] << '\n';
We will explain the standard-library algorithms much later (Chapter 21).
Once the temperatures are sorted, it’s easy to find the median: we just pick the middle element, the one with index temps.size()/2. If you feel like being picky (and if you do, you are starting to think like a programmer), you could observe that the value we found may not be a median according to the
definition we offered above. Exercise 3 at the end of this chapter is designed to solve that problem.
We deliberately left a potentially worse problem in that code. Did you spot it? What happens if we don’t enter any temperature values? In that case, all is well until we try to output the median temperature. Then, we try to read
temps[temps.size()/2] but temp.size() is zero, so there is no element to read!
Fortunately, if you use an implementation of vector that does range checking, the error is caught and reported. If you use our PPP support that’s what
happens. Forgetting the case of no elements is a common problem when iterating; always consider that possibility.