Answer:
Description: Write a MASM 32bit program with a loop and indexed addressing that calculates the sum of all thegaps between successive array elements. The array elements are doublewords, sequenced in nondecreasing order.
;Include Irvine32.inc file used with link library for 32 bit applications
.386
.model flat,stdcall
.stack 4096
ExitProcess proto,dwExitCode:dword
INCLUDE Irvine32.inc
.data
myArray DWORD 0,2,5,9,10
arrSize = ($-myArray)/TYPE myArray
gapArr DWORD arrSize-1 DUP(?)
sum DWORD ?
.code
main PROC
;Call the procedure
call Clrscr
;Initialize ESI pointer
mov esi, 0
mov ecx, arrSize
dec ecx
L1:
mov eax, myArray[esi+4]
sub eax, myArray[esi]
mov gapArr[esi], eax
inc esi
loop L1
;Calculate the sum of gaps between array elements
mov sum, 0
mov esi, 0
mov ecx, arrSize
dec ecx
; move gapArr[esi] to a temporary register and then add that register value to sum
L2:
mov edx, gapArr[esi]
add sum, edx
inc esi
loop L2
INVOKE ExitProcess,0
main ENDP
END main
Explanation:
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Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
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brainly.com/question/15051868
#SPJ4
