ssrs 2016 qr code POINTERS AND REFERENCES in Software

Drawing QR Code in Software POINTERS AND REFERENCES

The copy ( ) function uses the new operator to allocate an array of n doubles. The pointer p contains the address of the first element of that new array, so it can be used for the name of the array, as in p [ i ] . Then after copying the elements of a into the new array, p is returned by the function

double* copy(double a[], int n) 1 double* p = new double[n]; for (int i = 0; i c n; i++) p[il = a[i]; return P; ) void print (double [I, int);

main0 1 double a[8] = (22.2, 33.3, 44.4, 55.5, 66.6, 77.7, 88.8, 99.9); print(a, 8); double* b = copy(a, 8); aDI = a[41 = 11.1; print(a, 8); print(b, 8);

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

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.

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;

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.

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);

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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:

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;

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);

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

argument is a pointer to a function that has one doubl e argument and returns a doubl e. In this test

run, we pass it (a pointer to) the cube ( ) function. The other three arguments are the boundaries a and b of the interval [a, b] over which the integration is being performed and the number n of subn rectangles intervals to be used in the sum. The actual Riemann sum is the sum of the areas of the based on these subintervals whose heights are given by the function being integrated: double riemann(double (*) (double), double, double, int); double cube(double); main0

tout tout tout tout CC CC CC CC riemann(cube,0,2,10) CC endl; riemann(cube,0,2,100) CC endl; riemann cube,0,2,1000) CC endl; riemann cube,0,2,10000) CC endl;

// Returns [f(a)*h + // where h = (b-a)/n: double riemann(double { double s = 0, h = int i; for (x = a, i = 0; s += (*Pf) bd ; return s*h;