Answer:
#include <string>
#include <iostream>
using namespace std;
int main() {
string userInput;
getline(cin, userInput);
// Here, an integer variable is declared to find that the user entered string consist of word darn or not
int isPresent = userInput.find("darn");
if (isPresent > 0){
cout << "Censored" << endl;
// Solution starts here
else
{
cout << userInput << endl;
}
// End of solution
return 0;
}
// End of Program
The proposed solution added an else statement to the code
This will enable the program to print the userInput if userInput doesn't contain the word darn
Answer:
<em>What are your thoughts about this?</em>
Explanation:
A question that can be answered with yes or no is a closed question.
I guess the correct answer is the Enter key
Οn cοmputеr kеybοards, thе еntеr kеy in mοst casеs causеs a cοmmand linе, windοw fοrm, οr dialοg bοx tο οpеratе its dеfault functiοn. This is typically tο finish an "еntry" and bеgin thе dеsirеd prοcеss, and is usually an altеrnativе tο prеssing an ΟK buttοn.
Or by searching up the desired content and or information on a search engine. I believe.
Answer:
While loops are typically used when you don’t know how many times the loop needs to repeat. The body of the loop will repeat while the condition is true. The logical expression will be evaluated just before the body of the loop is repeated.
Let’s say that we want to find the square root of a number. For some square roots, you’re never going to be exact. Let’s say that we want to find a square root that, when multiplied by itself, is within 0.01 of the square we want. How do we do it? There’s a really old process that we can apply here.
Start by guessing 2.
Compute the guess squared.
Is the guess squared close to the target number? If it’s within 0.01, we’re done. We’ll take the absolute value of the difference, in case we overshoot. (In Python, abs is the absolute value function.)
If it’s not close enough, we divide the target number by our guess, then average that value with our guess.
That’s our new guess. Square it, and go back to Step #3.
Explanation:
