Write the following three functions: a) The function gets an array A of length n of ints, and a boolean predicate pred. It returns the smallest index i such that pred(A[i])==true. If no such element is not found, the function returns -1. int find(int* A, int n, bool (pred)(int)); b) The function gets an array A of length n of ints, and a function f. It applies f to each element of A. void map(int A, int n, int (f)(int)); c) The function gets an array A of length n of ints, and a function f. The function f gets 2 ints and works as follows: 1. Start with accumulator = A[0] 2. For i=1...length-1 compute accumulator=f(accumulator, A[i]) 3. Return accumulator For example, if f computes the sum of the two inputs, then reduce() will compute the sum of the entire array. int reduce(int A, int n, int (f)(int,int)); Test for the functions: // used for test Q3-find bool is_even(int x) { return x%2 == 0; } bool is_positive(int x) { return x>0; } void test_q3_find() { intA[6] = {-1,3,-6,5,2,7}; if (find(A, 6, is_even)==2 && find(A, 6, is_positive)==1) printf("Q3-find ok\n"); else printf("Q3-find ERROR\n"); } // used for test Q3-map int mult4(int x) { return x4;} void test_q3_map() { intA[6] = {-1,3,-6,5,2,7}; map(A, 6, mult4); if (A[0]==-4 && A[3]==20) printf("Q3-map ok\n"); else printf("Q3-map ERROR\n"); } // used for test Q3-reduce int sum(int x, int y) {return x+y;} int last(int x, int y) {return y;} void test_q3_reduce() { intA[6] = {-1,3,-6,5,2,7}; if (reduce(A, 6, sum)==10 && reduce(A, 6, last)==7) printf("Q3-reduce ok\n"); else printf("Q3-reduce ERROR\n"); }

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter15: Recursion

Section: Chapter Questions

Problem 6PE

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Question

Programming Language: C

Write the following three functions:

a) The function gets an array A of length n of ints, and a boolean predicate pred.

It returns the smallest index i such that pred(A[i])==true.

If no such element is not found, the function returns -1.

int find(int* A, int n, bool (*pred)(int));

b) The function gets an array A of length n of ints, and a function f.

It applies f to each element of A.

void map(int* A, int n, int (*f)(int));

c) The function gets an array A of length n of ints, and a function f. The function f gets 2 ints

and works as follows:

1. Start with accumulator = A[0]

2. For i=1...length-1 compute accumulator=f(accumulator, A[i])

3. Return accumulator

For example, if f computes the sum of the two inputs, then reduce() will compute the sum of the

entire array.

int reduce(int* A, int n, int (*f)(int,int));

Test for the functions:

// used for test Q3-find

bool is_even(int x) { return x%2 == 0; }

bool is_positive(int x) { return x>0; }

void test_q3_find() {

intA[6] = {-1,3,-6,5,2,7};

if (find(A, 6, is_even)==2 && find(A, 6, is_positive)==1)

printf("Q3-find ok\n");

else

printf("Q3-find ERROR\n");

}

// used for test Q3-map

int mult4(int x) { return x*4;}

void test_q3_map() {

intA[6] = {-1,3,-6,5,2,7};

map(A, 6, mult4);

if (A[0]==-4 && A[3]==20)

printf("Q3-map ok\n");

else

printf("Q3-map ERROR\n");

}

// used for test Q3-reduce

int sum(int x, int y) {return x+y;}

int last(int x, int y) {return y;}

void test_q3_reduce() {

intA[6] = {-1,3,-6,5,2,7};

if (reduce(A, 6, sum)==10 && reduce(A, 6, last)==7)

printf("Q3-reduce ok\n");

else

printf("Q3-reduce ERROR\n");

}

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