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RoseWind [281]
3 years ago
9

Use the following data definitions data myBytes BYTE 10h,20h,30h,40h myWords WORD 3 DUP(?),2000h myString BYTE "ABCDE" What will

be the value of EAX after each of the following instructions execute? mov eax,TYPE myBytes ; a. mov eax,LENGTHOF myBytes ; b. mov eax,SIZEOF myBytes ; c. mov eax,TYPE myWords ; d. mov eax,LENGTHOF myWords ; e. mov eax,SIZEOF myWords ; f. mov eax,SIZEOF myString ; g.
Computers and Technology
1 answer:
Elena-2011 [213]3 years ago
8 0

Answer:

<em>Given Data:</em>

<em>myBytes BYTE 10h, 20h, 30h, 40h </em>

<em>myWords WORD 3 DUP(?), 2000h </em>

<em>myString BYTE "ABCDE"</em>

<em />

Based on the data that we are given we can conclude that:

(a).     a. EAX = 1

         b. EAX = 4

         c. EAX = 4

         d. EAX = 2

         e. EAX = 4

         f. EAX = 8

         g. EAX = 5

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The density of mercury is 13.6 grams per cubic centimeter. Complete the steps for converting 13.6 g/cm3 to kg/m3.
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7 0
2 years ago
Input 10 integers and display the following:
LekaFEV [45]

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

3 0
3 years ago
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