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## Checking the Other Values

howMany only counts the occurrences of a particular value, and we are interested in seeing how many times each value appears. We can solve that problem with a loop:

int numValues = 20;
int upperBound = 10;
pvector<int> vector = randomVector (numValues, upperBound);

cout << "value\thowMany";

for (int i = 0; i<upperBound; i++) {
cout << i << '\t' << howMany (vector, i) << endl;
}

Notice that it is legal to declare a variable inside a for statement. This syntax is sometimes convenient, but you should be aware that a variable declared inside a loop only exists inside the loop. If you try to refer to i later, you will get a compiler error.

This code uses the loop variable as an argument to howMany, in order to check each value between 0 and 9, in order. The result is:

value   howMany
0       2
1       1
2       3
3       3
4       0
5       2
6       5
7       2
8       0
9       2

Again, it is hard to tell if the digits are really appearing equally often. If we increase numValues to 100,000 we get the following:

value   howMany
0       10130
1       10072
2       9990
3       9842
4       10174
5       9930
6       10059
7       9954
8       9891
9       9958

In each case, the number of appearances is within about 1% of the expected value (10,000), so we conclude that the random numbers are probably uniform.

Last Update: 2005-12-05