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:
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.
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Answer:
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Answer:
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Answer:
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Answers
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Answer:
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I can't understand. Please tell clearly.
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Answers
Answer is given below
Explanation:
differences between a file-processing system and a database management system is given below
file-processing system
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Answer:
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The variable, A, starts with the value, 1 The variable, B, starts with the value, 10 The variable, C, starts with the value, 100. The variable, X, starts with the value, 0 Store the value of C times 3 in X. Add the value of B times 6 to the value already in X Add the value of A times 5 to the value already in X. Display the value in X on the screen. Output #3 Pseudocode #3
Answers
Answer:
1. Start
2. A = 1
3. B = 10
4. C = 100
5. X = 0
6. X = C * 3
7. X = X + B * 6
8. X = X + A * 5
9. Print X
10. Stop
Explanation:
Required
Write the Pseudocode for the instruction
The pseudocode begins and ends with a start and stop statement
The next 4 lines initializes A, B, C and X
2. A = 1
3. B = 10
4. C = 100
5. X = 0
The next 3 lines calculates the new value of X base on the given instruction
6. X = C * 3
7. X = X + B * 6
8. X = X + A * 5
This line displays the value of X to the screen
9. Print X
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gdghfhffhgcgchchghvb. bvhvhhvhvhgcgfhsufsjtstjsfsfshrs
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Answers
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Explanation:
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Answers
Answer:
I need to send black-and-white images in a print-ready format. Therefore, here are the steps I’d follow:
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2. I’d use Photoshop to modify or correct the color of the images.
3. I’d convert the images into black-and-white versions.
4. I’d save the selected files in the TIFF format. This method will ensure that the images retain their quality and are print-ready.
5. I’d also convert these files into the JPEG format to get good-quality, low-resolution images for the client’s web publishing.
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I’d send all these saved copies to the magazine for their work.
Explanation:
This is the answer straight from Plato. I hope that this was helpful to you. Please like and rate my answer if it was! :)
Write a JAVA CODE recursive function that when passed a value n displaysn (n-1) (n-2) (n-3) ... 0 ... (n-3) (n-2) (n-1) nfor example, if passed 5 displays 5 4 3 2 1 0 1 2 3 4 5
Answers
//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.