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patriot [66]
3 years ago
15

Create a function called makePositive that accepts a single argument: an integer array. The function should walk through the arr

ay and change any negative numbers to positive. You can assume that the array passed to the function will have a 0 at the end of the array. In other words, you do not know the size of the array; you instead know that a certain condition will be true at the end of the array.
Computers and Technology
1 answer:
coldgirl [10]3 years ago
8 0

Answer:

This solution is implemented in C++

void makePositive(int arr[]){

   int i =0;

   while(arr[i]!=0){

       if(arr[i]<0){

           arr[i] = abs(arr[i]);

       }

       i++;

   }

   i = 0;

   while(arr[i]!=0){

    cout<<arr[i]<<" ";

    i++;

}  

}

Explanation:

This defines the function makePositive

void makePositive(int arr[]){

This declares and initializes i to 0

   int i =0;

The following iteration is repeated until the last element in the array

   while(arr[i]!=0){

This checks if current array element is negative

       if(arr[i]<0){

If yes, it changes it to positive

           arr[i] = abs(arr[i]);

       }

The next element is then selected

       i++;

   }

This sets i to 0

   i = 0;

The following iteration prints the updated content of the array

<em>    while(arr[i]!=0){ </em>

<em>     cout<<arr[i]<<" "; </em>

<em>     i++; </em>

<em> }   </em>

}

See attachment for full program which includes the main

Download cpp
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Write a loop that reads positive integers from standard input and that terminates when it reads an integer that is not positive.
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Answer:

The program to this question as follows:

Program:

#defining variable and assign value

odd=0  

even=0  

i=1  

print('Input 0 after inserting all number: ') #print message

while i>0: #define loop for check inserting and add all number

   i=int(input()) #input value by user end

   if (i< 0): #check value is not positive

       break

   if ((i % 2)==0and(i>0)): #check even number condition

       even=even+i #add even numbers

   if ((i % 2)!=0and(i>0)): #check odd number condition

       odd=odd+i #add odd numbers

print ('Sum of Even number:',even,'and','Sum of Odd number:',odd) #print value

Output:

Input 0 after inserting all number:  

1

2

3

5

7

0

Sum of Even number: 2 and Sum of Odd number: 16

 Explanation:

In the above python program, three variable is defined, that is "even, odd and i", in which "even and odd" variable assign value that is "0", and the variable i is used for inserting elements from user. In the next line a while loop is defined that inserts numbers in variable "i", in this loop, 3 if block is used that can be defined as follows:

  • In first if block variable "i", checks that value is non-positive number if this condition is true, it will break the loop.
  • In second if block this block check even number condition and add in even variable.
  • In third, if block this block check odd number condition and add in odd variable.  

End of the loop we use print function that prints the sum of even numbers and odd numbers.

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