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Nataliya [291]
3 years ago
15

Write a C program that reads two hexadecimal values from the keyboard and then stores the two values into two variables of type

unsigned char. Read two int values p and k from the keyboard, where the values are less than 8. Replace the n bits of the first variable starting at position p with the last n bits of the second variable. The rest of the bits of the first variable remain unchanged. Display the resulting value of the first variable using printf %x.Test Cases:n = 3;p=4;input: a= 0x1f (0001 1111) b = c3 (1100 0011); output: f (0000 1111)n = 2;p=5;input: a= 0x1f (0001 1111) b = c3 (1100 0011); output: 3f (0011 1111)
Computers and Technology
1 answer:
sattari [20]3 years ago
4 0

Solution :

#include  $$

#include $$

#include $$

//Converts $\text{hex string}$ to binary string.

$\text{char}$ * hexadecimal$\text{To}$Binary(char* hexdec)

{

 

long $\text{int i}$ = 0;

char *string = $(\text{char}^ *) \ \text{malloc}$(sizeof(char) * 9);

while (hexdec[i]) {

//Simply assign binary string for each hex char.

switch (hexdec[i]) {

$\text{case '0'}:$

strcat(string, "0000");

break;

$\text{case '1'}:$

strcat(string, "0001");

break;

$\text{case '2'}:$

strcat(string, "0010");

break;

$\text{case '3'}:$

strcat(string, "0011");

break;

$\text{case '4'}:$

strcat(string, "0100");

break;

$\text{case '5'}:$

strcat(string, "0101");

break;

$\text{case '6'}:$

strcat(string, "0110");

break;

$\text{case '7'}:$

strcat(string, "0111");

break;

$\text{case '8'}:$

strcat(string, "1000");

break;

$\text{case '9'}:$

strcat(string, "1001");

break;

case 'A':

case 'a':

strcat(string, "1010");

break;

case 'B':

case 'b':

strcat(string, "1011");

break;

case 'C':

case 'c':

strcat(string, "1100");

break;

case 'D':

case 'd':

strcat(string, "1101");

break;

case 'E':

case 'e':

strcat(string, "1110");

break;

case 'F':

case 'f':

strcat(string, "1111");

break;

default:

printf("\nInvalid hexadecimal digit %c",

hexdec[i]);

string="-1" ;

}

i++;

}

return string;

}

 

int main()

{ //Take 2 strings

char *str1 =hexadecimalToBinary("FA") ;

char *str2 =hexadecimalToBinary("12") ;

//Input 2 numbers p and n.

int p,n;

scanf("%d",&p);

scanf("%d",&n);

//keep j as length of str2

int j=strlen(str2),i;

//Now replace n digits after p of str1

for(i=0;i<n;i++){

str1[p+i]=str2[j-1-i];

}

//Now, i have used c library strtol

long ans = strtol(str1, NULL, 2);

//print result.

printf("%lx",ans);

return 0;

}

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Answer:

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Explanation:

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If the driver is not scanning the road, then DIC will help the driver to find lots of useful information which keeps himself as well as the peer drivers on the road safe and thus avoiding accidents where by saving lives and heavy injury.

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Explanation:

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Write a function DrivingCost() with input parameters drivenMiles, milesPerGallon, and dollarsPerGallon, that returns the dollar
posledela

Answer:

#include <iostream>

using namespace std;

double DrivingCost(int drivenMiles,double milesPerGallon,double dollarsPerGallon)

{

   double dollarsperMile=dollarsPerGallon/milesPerGallon;//calculating dollarsperMile.

   return dollarsperMile*drivenMiles;//returning thr driving cost..

}

int main() {

double ans;

int miles;

cout<<"Enter miles"<<endl;

cin>>miles;

ans=DrivingCost(miles,20.0,3.1599);

cout<<ans<<endl;

return 0;

}

Output:-

Enter miles

10

1.57995

Enter miles

50

7.89975

Enter miles

100

15.7995

Explanation:

In the function first I have calculated the dollars per mile and after that I have returned the product of dollarspermile and driven miles.This will give the cost of the Driving.

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Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.
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Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

v = \frac{(xf-xo)}{(t-to)}

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

t =\frac{xf}{v}  =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.

⇒ t = 8.7*10⁻⁴ sec.

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3 years ago
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