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function to examine the In this run we initialize a as an array of 8 doubles. We use a print ( ) contents of a. The the copy ( ) function is called and its return value is assigned to the pointer b which then serves as the name of the new array. Before printing b, we change the values of two of a's elements in order to check that b is not the same array as a, as the last two print ( ) calls confirm. 6.28

Fig. 5-29

Write a function that uses pointers to search for the address of a given integer in a given array. If the given integer is found, the function returns its address; otherwise it returns NULL.

(5.5)

We use a for loop to traverse the array. If the target is found at a [ i ] , then its address &a [ i ] is returned. Otherwise, NULL is returned: int* location(int a[], int n, int target) { for (int i = 0; i < n; i++) if (a[i] == target) return &a[i]; return NULL;

Fig. 5-30

Another useful feature is the ability to deactivate a menu item by deselecting the Enabled box (see Fig. 5.3). The menu item will then appear grayed out within the menu, and it will not respond to mouse clicks, keyboard access characters or keyboard shortcuts. In addition, a menu item may be made invisible (and inactive) by deselecting the Visible box (see Fig. 5.3). Both of these features can later be changed under program control. Finally, the items within a menu can be grouped together by introducing separators at various locations within a menu. Each separator is a menu item consisting only of a single dash (minus sign). Note that each separator must follow the same rules of indentation as its surrounding menu items.

The test driver calls the function and stores its return address in the pointer p. If that is nonzero (i.e., not NULL), then it and the int to which it points are printed.

Assuming ABCD in Fig. 5-31 is a parallelogram, find x and y if (a) AD (b) AB (c) m/A (d) m/A 5x, AB 2x, BC 4y 2x, CD 3y y, perimeter 7x 84 5y 10

main0 I { int a[81 = (22, 33, 44, 55, 66, 77, 88, 99}, * p, n; do { tin >> n; \ if (p = location(a, 8, n)) tout CC p CC ", ' CC *p -C-C endl; else tout CC n CC ' was not found.\n"; } while (n > 0);

(5.3)

0:. 0: 6.29 _ .

__,. was. :.

Fig. 5-31

.,.. fOUti@.,

Fig. 5-32

,: . _.

EXAMPLE 5.4 MORE MENU ENHANCEMENTS (GEOGRAPHY REVISITED)

Assuming ABCD in Fig. 5-32 is a parallelogram, find x and y if (a) AE (b) AE (c) AE (d) AE x y, EC 20, BE x x y, ED 9 7, ED 24 x y 8

1 .

(5.3)

Write a function that is passed an array of n pointers to floats and returns a newly created array that contains those n float values.

We use a for loop to traverse the array until p points to the target:

Provide the proofs requested in Fig. 5-33. (5.6)

float* duplicate(float* p[], int n) -I float* const b = new float[n]; for (int i = 0; i c n; i++, q++) b[i] = *p[i]; return b;

Fig. 5-33

float a[81 = (44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 55.5); print(a, 8); float* p[8]; for (int i = 0; i c 8; i++) // p[i] points to a[i] pE-1 = &a[i]; print@, 8); float* const b = duplicate(p, 8); print(b, 8);

Prove each of the following: (a) The opposite sides of a parallelogram are congruent (Principle 3). (b) If the opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram (Principle 8). (c) If two sides of a quadrilateral are congruent and parallel, the quadrilateral is a parallelogram (Principle 9). (d) The diagonals of a parallelogram bisect each other (Principle 6). (e) If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram (Principle 11).

This function, named riemann ( ) , is similar to the sum ( )