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7.1 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) { double* p = new double[n]; for (int i = 0; i < n; i++) p[i] = a[i]; return p; } void print(double [], int); int main() { 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); a[2] = a[4] = 11.1; print(a, 8); print(b, 8); } 22.2, 33.3, 44.4, 55.5, 66.6, 77.7, 88.8, 99.9 22.2, 33.3, 11.1, 55.5, 11.1, 77.7, 88.8, 99.9 22.2, 33.3, 44.4, 55.5, 66.6, 77.7, 88.8, 99.9 In this run we initialize a as an array of 8 doubles. We use a print() function to examine the contents of a. 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. 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. int main() { int a[8] = {22, 33, 44, 55, 66, 77, 88, 99}, * p, n; do { cin >> n; if (p = location(a, 8, n)) cout << p << ", " << *p << endl; else cout << n << " was not found.\n"; } while (n > 0); }
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44 0x3fffcc4, 44 50 50 was not found. 99 0x3fffcd8, 99 90 90 was not found. 0 0 was not found. We use a for loop to traverse the array until p points to the target: float* duplicate(float* p[], int n) { float* const b = new float[n]; for (int i = 0; i < n; i++) b[i] = *p[i]; return b; } void print(float [], int); void print(float* [], int); int main() { float a[8] = {44.4, 77.7, 22.2, 88.8, 66.6, print(a, 8); float* p[8]; for (int i = 0; i < 8; i++) p[i] = &a[i]; // p[i] points to a[i] print(p, 8); float* const b = duplicate(p, 8); print(b, 8); } 44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9,
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This function, named riemann(), is similar to the sum() function in Example 7.18. Its first argument is a pointer to a function that has one double argument and returns a double. 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 subintervals to be used in the sum. The actual Riemann sum is the sum of the areas of the n rectangles based on these subintervals whose heights are given by the function being integrated: double riemann(double (*)(double), double, double, int); double cube(double); int main() { cout << riemann(cube,0,2,10) << endl; cout << riemann(cube,0,2,100) << endl; cout << riemann(cube,0,2,1000) << endl; cout << riemann(cube,0,2,10000) << endl; } // // Returns [f(a)*h + f(a+h)*h + f(a+2h)*h + . . . + f(b-h)*h], where h = (b-a)/n:
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