A ___ block unauthorized access to a network or an individual computer

* Declare a variablecalled "car" of type "Car", and initialise its value to a new instance of the "Car"class.

kicyunya [14]

Answer:

Car car = new Car();

Explanation:

Here we declare a variable called car which is a reference to an object of type Car.

Next we create a new object of type Car by using the new keyword. Here we are invoking no argument constructor of class Car. In case the car class has other constructors which take additional arguments, we could have initialized our object using the corresponding version of new.

The reference variable car is initialized then to the newly created object of type Car as described in the previous step. Now we can use the car reference variable to invoke relevant methods defined in the Car class.

3 0

2 years ago

At the frequency of 2.4 GHz what is the free-space path loss in dB.

chubhunter [2.5K]

Answer:

The FSBL is 100.05 dB

Solution:

As per the question:

Frequency, f = 2.4 GHz = $2.4\times 10^{9}$

Now, we know that to calculate FSBL, i.e., Free Space Path Loss in dB, we use:

$FSBL(in dB) = 10log_{10}(\frac{4\pifd}{c})^{2}$

where

d = 1 km = 1000

c = $3\times 10^{8} m/s$

$FSBL(in dB) = 20log_{10}\frac{4\pifd}{c} = 20log_{10}(fd) + 20log_{10}(\frac{4\pi}{c})$

$FSBL(in dB) = 20log_{10}(f) + 20log_{10}(d) + 20log_{10}(\frac{4\pi}{c})$

$FSBL(in dB) = 20log_{10}(2.4\times 10^{9}) + 20log_{10}(10^{3}) + 20log_{10}(\frac{4\pi}{3\times 10^{8}})$

$FSBL(in dB) = 20\times 9.3802 + 20\times 3 - 147.55 = 100.05 dB$

6 0

2 years ago

How to write a function that counts the letters in a string in C?

stiv31 [10]

Answer:

Here is the C function that counts the letters in a string:

int LetterCount(char string[]){ //function to count letters in a string passed as parameter

string[100]; // char type string with size 100

int i, letters; // letter variable stores the count for no. of letters in string

i = letters = 0; //initialize variables i and letters to 0

while (string[i] != '\0') { // loop iterates through the entire string until end of string is reached

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') ) { // if condition checks the letters in the string

letters++; } // increments 1 to the count of letters variable each time a letter is found in the string

i++; } //increments value of i to move one character forward in string

printf("Number of Letters in this String = %d", letters); // displays the number of letters in the string

return 0; }

Explanation:

Here the question means that the function should count the letters in a string. So the letters are the alphabets from a to z and A to Z.

Here is the complete program:

#include <stdio.h> // to use input output functions

int LetterCount(char string[]){ // method that takes a character string as parameter and counts the letters in the string

string[100];

int i, letters;

i = letters = 0;

while (string[i] != '\0') {

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z'))

{letters++; }

i++; }

printf("Number of Alphabets in this String = %d", letters);

return 0;}

int main(){ // start of main function

char string[100]; //declares a char array of string

printf("Please Enter a String : "); //prompts user to enter a string

fgets(string,100,stdin); //get the input string from user

LetterCount(string); } // calls method to count letters in input string

I will explain this function with an example

Lets suppose the user input "abc3" string

string[100] = "abc3"

Now the function has a while loop that while (string[i] != '\0') that checks if string character at i-th position is not '\0' which represents the end of the character string. As value of i = 0 so this means i is positioned at the first character of array i.e. 'a'

At first iteration:

i = 0

letter = 0

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') ) if condition checks if the character at i-th position of string is a letter. As the character at 0-th position of string is 'a' which is a letter so this condition evaluates to true. So the statement letter++ inside if condition executes which increments the letter variable to 1. So the value of letter becomes 1. Next statement i++ increments the value of i to 1. So i becomes 1. Hence:

i = 1

letter = 1

At second iteration:

i = 1

letter = 1

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 1st position of string is 'b' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 2

letter = 2

At third iteration:

i = 2

letter = 2

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 2nd position of string is 'c' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 3

letter = 3

At fourth iteration:

i = 3

letter = 3

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 3rd position of string is '3' which is not a letter but a digit so this condition evaluates to false. So the statement letter++ inside if condition does not execute. Now i++ executes which increments this variable to 1. Hence:

i = 4

letter = 3

Now the loop breaks because while (string[i] != '\0') condition valuates to false as it reaches the end of the string.

So the statement: printf("\n Number of Letters in this String = %d", letters); executes which prints the value of letters on the output screen. Hence the output is:

Number of Letters in this String = 3