Section 10.11 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;
vector<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.

Checkpoint 10.11.1.

If you declare a variable inside a for statement, where can it exist?

inside of the for loop.
Correct!
outside of the for loop, but inside of the function it’s used in.
Incorrect! The variable goes out of scope as soon as the for loop terminates!
outside of the function, and everywhere else in the program.
Incorrect! The variable goes out of scope as soon as the for loop terminates!

Checkpoint 10.11.2.

Multiple Response When we increase the size of numValues, which of the following is true:

the difference between actual and expected number of appearances increases
Correct! The numbers go from being off by less than 5 to more than 100.
the difference between actual and expected number of appearances decreases
Incorrect! Take a look at the numbers again!
the percent by which the number of appearances differs from the expected number increases
Incorrect! Take a look at the numbers again!
the percent by which the number of appearances differs from the expected number decreases
Incorrect! As we continue to increase the size of numValues, the percent by which the number of appearances differes from the expected value approaches 0.

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