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The printQuizAverages() function prints the average of each of the 3 rows of scores, while the printClassAverages function prints the average of each of the 5 columns of scores. Here is an interactive of the program:
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EXAMPLE 5.19 Processing a Three-Dimensional Array This program simply counts the number of zeros in a three-dimensional array: int numZeros(int a[][4][3], main0 1 int a[2][4][3] int nl, int n2, int n3);
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= 1 -I 15,0,2L uLO,9L WJ,OL 17,717) 1, 1 {3,O,OL -M,LOL W,O,OL C&O,91 1 ); tout << "This array has ' << numZeros(a,2,4,3) << ' zeros:\n";
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int numZeros(int a[][4][3], int nl, int n2, int n3) { int count = 0; for (int i = 0; i c nl; i++) for (int j = 0; j c n2; j++) for (int k = 0;. k < n3; k++) if (a[i][j][k] == 0) ++count; return count; 1 Notice how the array is initialized: it is a 2-element array of 4-element arrays of 3 elements each. That makes a total of 24 elements. It could have been initialized like this: int a[2][4][3] = ~~,0,2,0,0,9,4,~,0,7,7,7,3,0,0,8,5,o,o,o,~,~,~,~~; or like this: int a[21 [4][3] = 1 ~5,0,2,0,0,9,4,l,O,7,7,7), ~3,0,0,8,~,0,0,0,0,2,0,9) ); But these are more difficult to read and understand than the three-dimensional initializer list. Also notice the three nested for loops. In general, processing a d-dimensional array is done with d for loops, one for each dimension.
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How many different types can the elements of an array have What type and range must an array s subscript have What values will the elements of an array have when it is declared if it does not include an initializer What values will the elements of an array have when it is declared if it has an initializer with fewer values than the number of elements in the array What happens if an array s initializer has more values than the size of the array How does an enum statement differ from a typedef statement When a multi-dimensional array is passed to a function, why does C++ require all but the first dimension to be specified in the parameter list 0
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Write and run a program that reads an unspecified number of numbers and then prints them together with their deviations from their mean.
We can accumulate the numbers as they are read in and then compute the mean (average) just by dividing their sum by their count: const int size = 100; main0 Terminate with O:\n"; tout CC "Enter data. double a[size] I xr sum = 0.0; for (int n = 0 ; ; n++) 1 tin >> x; if (x == 0 ) break; a[n] = x; sum += x; double mean = sum/n; mean CC endl; tout CC "mean for (int i = 0; i c n; i++) cc a[i] - mean CC endl; tout cc CC a[i] cc
= II << I I I I \t \t
The input loop continues until 0 is read. The, deviations are printed as a [ i
- mean.
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Write and test the function
void insert(int a[], int& n, int x)
This function inserts the item x into the sorted array a of n elements and increments n. The new item is inserted at the location that maintains the sorted order of the array. This requires shifting elements forward to make room for the new X.
Our test driver defines an array of size 100 and initializes it with 10 elements in increasing order: void print(int [I, int); void insert(int [I, int&, int); main0 int a[1001 = { 261, 288, 289, 301, 329, int n = 10, x; print(a, n); tout << "Item to be inserted: "; tin >> x; insert(a, n, x); print(a, n); 333, 345, 346, 346, 350);
void print(int a[], int n) 1 for (int i = 0; i < n-l; i++) { tout CC a[i] << 'I, 1'; if ((i+1)%16 == 0) tout CC endl; > tout CC a[n-1] cc endl;
void insert(int a[], int&
n, int x)
for (int i = n; i > 0 SC& a[i-l] > x; i--) a[il = a[i-11; a[i] = x; ++n;
T h e insert ( ) function works from the high end of the array, searching backward for the correct location to put x. As it searches, it shifts the elements that are larger than x one place to the right to make way for x. On the first run, 300 is inserted by shifting 7 elements to the right.
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