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3 years ago

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Computers and Technology

1 answer:

max2010maxim [7]3 years ago

5 0

Answer:

its .doc

Explanation:

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Answer:

1/3

Explanation:

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2 years ago

Given a Student class, create a class with following characteristics:

Kay [80]

Answer:

ssume that,

Maximum “students” count can be 10.

*Driver Class*

*Solution class*

import java.util.*;

class Student {

private String name;

private int rollNo;

public String getName() {}

public void setName(String name) {}

public int getRollNo() {}

public void setRollNo(int rollNo) {}

};

class ClassRoom {

private int i;

private Student[] students;

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public Student[] getAllStudents() {}

};

8 0

3 years ago

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

sattari [20]

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;

}

4 0

3 years ago