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mariarad [96]
3 years ago
7

0001 0000 1100 0011 (unsigned, that is, interpreted as b2u)i.8 bit truncated value:ii.does the value change when truncated to 8

bits?iii.how did you decide?d.1101 1111 0011 1010 (signed, that is, interpreted as b2t)i.8 bit truncated value:ii.does the value change when truncated to 8 bits?iii.how did you decide?
Computers and Technology
1 answer:
mart [117]3 years ago
4 0
I don't know what b2u and b2t are but I'll manage.

0001 0000 1100 0011 unsigned

i. We'll assume when we' truncate we get the low order bits:

1100 0011 unsigned

ii. Those are different

iii.  There's a 1 in the high order bits that got thrown away when we truncated.

1101 1111 0011 1010 <span>signed </span>

i. Truncated:

0011 1010 

Typically when we'd coerce a signed twos complement value to eight bits we'd preserve the sign.  Calling it truncation makes it sound to me that wasn't intended.

ii. yes, different value

iii. Again, we threw out bits


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Output a table that show the cube of the numbers 1-15<br> (C++)
Rainbow [258]

Answer:

The c++ program to display cube of numbers from 1 to 15 is given below.

#include <iostream>

using namespace std;

int main() {    

   // variables declared and initialized  

   int num = 15, k, cube;    

   cout << " Cubes of numbers from 1 to 15 are shown below " << endl;    

   for( k = 1; k <= num; k++ )

   {

       // cube is calculated for each value of k and displayed

       cube = k * k * k ;

       cout << " \t " << cube << endl;

   }

return 0;

}

 

OUTPUT

Cubes of numbers from 1 to 15 are shown below  

  1

  8

  27

  64

  125

  216

  343

  512

  729

  1000

  1331

  1728

  2197

  2744

  3375

Explanation:

The variables are declared and initialized for loop, cube and for condition in the loop – k, cube, num respectively.

Inside for loop which executes over k, beginning from 1 to 15, cube of each value of k is calculated and displayed. The loop executes 15 times. Hence, cube for numbers from 1 to 15 is displayed after it is calculated.

   for( k = 1; k <= num; k++ )

   {

      cube = k * k * k ;

       cout << " \t " << cube << endl;

   }

In the first execution of the loop, k is initialized to 1 and variable cube contains cube of 1. Hence, 1 is displayed on the screen.

In the second execution, k is incremented by 1 and holds the value 2. The variable cube contains cube of 2, which is calculated, and 8 gets displayed on the screen.

After each execution of the loop, value of k is incremented.

After 15 executions, value of k is incremented by 1 and k holds the value 16. This new value is greater than num. Hence, the loop terminates.

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Answer:

The code is given as follows,

Explanation:

Code:

#include <stdio.h>

#include <string.h>  

int n; //to store size of array  

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{

  int i, j;

  int sum = 0;

  for(i = 0; i < n; i++)

  {

      sum = sum + arr[i]; //count sum

  }  

  for(i = 0; i < n; i++)

  {

      for(j = i+1; j < n; j++) //loop for every two elements in array

      {

          if(arr[i]*arr[j] > 2*sum) //check if proudct of two elements > 2 times sum

          {

              printf("\n%d x %d = %d, 2xSum = %d\n", arr[i], arr[j], arr[i]*arr[j], 2*sum);

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          }

      }

  }

  return "false"; // If proudct of two elements < 2 times sum. return false

}  

int main()

{  

  printf("\nEnter size of array: ");

  scanf("%d", &n); //read size of array

  int A[n]; //array of size n

  printf("\nEnter array elements: ");

  int i;

  for(i = 0; i < n; i++)

  {

      scanf("%d", &A[i]); //read array from stdin

  }

  printf("%s\n",ArrayChallenge(A)); //ccall function and print answer

 

  return 0;

}

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