Question

1. (40 pts) Consider the following function. Complete the function that does the following: (a) Take three arguments, int array, int as the length of the array, and int *max. (b) Find the maximum value and the minimum value in the array. (c) The pointer max points to the maximum value. (d) Return the pointer that points to the minimum value max)( findMinAndMax(int al l, int length, int int maxValue a[0], minValue al0 int int maxldx 0, minldx 0; for (int idx -1; idx< length; idx+)( if (aidx]> maxValue) maxvalue-alidxj maxldx idx; if (a[idx< minValue) minValue alidx] minldx - idx; 2. (60 pts) Consider the above function. Declare an integer array called myArray of length 5 Initialize myArray to 4, 1,-5, 3, 5, in sequence. Declare two pointers ptrMin and ptrMax, both pointing to integer. Call the above function findMinAndMax so that the pointer ptrMin points the the minimum value in the array and the pointer ptrMax points to the maximum value in the array

Answer 1

Solution:

The min and max is printed with the help of driver function, please have a look.

code:

#include<iostream> //i/o library

using namespace std;

int *findMinAndMax(int a[], int length, int *max)

{

int maxValue= a[0], minValue= a[0];

int maxIdx=0, minIdx=0;

for (int idx = 1; idx<= length; idx++)

{

if (a[idx] > maxValue) {

maxValue= a[idx];  

maxIdx = idx;

}

if (a[idx]<minValue) {

minValue = a[idx];

minIdx= idx;

}

}

*max= a[maxIdx];

return &minValue;

}

int main(){

int myArray[]= {4, 1, -5, 3, 5};

int max, *min;

min= findMinAndMax(myArray, 5, &max);

cout<<*min<<" "<<max;

return 0; //This is to make sure that the program terminates

}

Output:

FStudy\ Chegg 2k181codes\C-C++findMinMax.exe Process exited after 0.06464 seconds with return value Press any key to continue.

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)