Question

Consider the array called myArray declared below that contains positive V and negative integers. Write a loop statement to count and display the number of positive integers in the array. That is, your loop should display the message "Number of positive integers is 5". int myArray[]={-1,3,-9,9,33,-4,-5,100,4,-23};

Answer 1

Code: it works even if the elements in the array are changed

#include<stdio.h>
main()
{
// Array given in the question
int myArray[] = {-1,3,-9,9,33,-4,-5,100,4,-23};
// determine the length of the array
/* dividing total number of bytes occupied by array
with number of bytes per element gives number of elements*/
int len = sizeof(myArray)/sizeof(myArray[0]);
// initialize count to 0
int i, count = 0;
// loop over all the elements of array
for(i = 0; i<len; i++)
{
// if current element of array is greater than or equal to 0,
// increment count
if(myArray[i] >= 0)
count++;
}
// print the value of count
printf("Number of positive integers is %d\n", count);
}

Screenshot of code:

sample.c X # include<stdio. h> main () 2 // Array given in the question int myArray [ ] {-1,3,-9, 9, 33,-4,-5, 100,4,-23); / determine the length of the array /dividing total number of bytes occupied by array with number of bytes per element gives number of elements/ int len sizeof (myArray) /sizeof (myArray[0]); //initialize count to 0 int i, count 0: / loop over all the elements of array for (i= 0; i< len ; i++) 7 10 12 13 14 15 16 17 18 19 20 21 / af current element of array is greater than or equal to o // increment count i f (myArray [ i ] >-0)! count++ // print the value of count printf("Number of positive integers is %d\n", count);
Output:

Laptop\Chegg filessample.exe Number of positive integers is 5 NLaptop Chegg files)