C++

Set hasDigit to true if the 3-character passCode contains a digit.

#include
#include
#include
using namespace std;

int main() {
bool hasDigit;
string passCode;

hasDigit = false;
cin >> passCode;

/* Your solution goes here */

if (hasDigit) {
cout << "Has a digit." << endl;
}
else {
cout << "Has no digit." << endl;
}

return 0;

Answers

Answer 1

Answer:

Add this code the the /* Your solution goes here */  part of program:

for (int i=0; i<3; i++) { //iterates through the 3-character passCode

  if (isdigit(passCode[i])) //uses isdigit() method to check if character is a digit

           hasDigit = true;    } //sets the value of hasDigit to true when the above if condition evaluates to true

Explanation:

Here is the complete program:

#include <iostream> //to use input output functions

using namespace std; // to identify objects like cin cout

int main() { // start of main function

bool hasDigit; // declares a bool type variable  

string passCode; //declares a string type variable to store 3-character passcode

hasDigit = false; // sets the value of hasDigit as false initially

cin >> passCode; // reads the pass code from user

for (int i=0; i<3; i++) { //iterate through the 3 character pass code

   if (isdigit(passCode[i])) // checks if any character of the 3-character passcode contains a digit

     hasDigit = true;    }      //sets the value of hasDigit to true if the passcode contains a digit    

if (hasDigit) { // if pass code has a digit

cout << "Has a digit." << endl;} //displays this message when passcode has a digit

else { //if pass code does not have a digit

cout << "Has no digit." << endl;} //displays this message when passcode does not have a digit

return 0;}

I will explain the program with an example. Lets say the user enters ab1 as passcode. Then the for loop works as follows:

At first iteration:

i = 0

i<3 is true because i=0

if (isdigit(passCode[i]) this if statement has a method isdigit which is passed the i-th character of passCode to check if that character is a digit. This condition evaluates to false because passCode[0] points to the first character of pass code i.e. a which is not a digit. So the value of i is incremented to 1

At second iteration:

i = 1

i<3 is true because i=1

if (isdigit(passCode[i]) this if statement has a method isdigit which is passed the i-th character of passCode to check if that character is a digit. This condition evaluates to false because passCode[1] points to the second character of pass code i.e. b which is not a digit. So the value of i is incremented to 1

At third iteration:

i = 2

i<3 is true because i=2

if (isdigit(passCode[i]) this if statement has a method isdigit which is passed the i-th character of passCode to check if that character is a digit. This condition evaluates to true because passCode[3] points to the third character of pass code i.e. 1 which is a digit. So the hasDigit = true;  statement executes which set hasDigit to true.

Next, the loop breaks at i=3 because value of i is incremented to 1 and the condition i<3 becomes false.

Now the statement if (hasDigit) executes which checks if hasDigit holds. So the value of hasDigit is true hence the output of the program is:

Has a digit.

C++ Set HasDigit To True If The 3-character PassCode Contains A Digit. #include #include #include Using

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Answer:

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Answer:

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Answer: genetic algorithm


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Answers

Answer:

//Begin class declaration

public class RecursiveFunction {

   

   //Write function

   //It takes in an int value as parameter

   //And returns a string value.

  public static String recursivefn(int n){

       //Check the value of variable n

       //If the value is less than or equal to 0

       //Return the string "0"

       if(n <= 0){

           return "0";

       }

       

       //If n is not less than or equal to zero,

       //return a string that has:

       //1. n,

       //2. followed by a space,

       //3. followed by another call to the function with n - 1

       //4. followed by a space,

       //5. followed by n

       return n + " " + recursivefn(n-1) + " " + n;

   }

   

   //Begin main method

    public static void main(String []args){        

       //Call the recursivefn() on 5

       System.out.println(recursivefn(5));        

    } //End of main method

   

} //End of class declaration

Sample Output:

5 4 3 2 1 0 1 2 3 4 5

Explanation:

The code has been written in Java. It contains comments explaining important parts of the code. Please go through the comments.

A sample output has also been provided.

The program file and a screenshot of the output have also been attached to this response.

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Answers

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Answers

Answer:

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