int input; while (true){ scanf(“%d”,&input); if (input>=1 && input<=10){ // process with your input then use break to end the while loop } else{ printf(“Wrong input! try Again.”); continue; } } 1 solved How to prompt the user to enter a integer within a certain amount of numbers
result and counter are separate variables with different goals in this code. counter is incremented like counter += 1 so that eventually the while condition while (counter<10) will be satisfied and the code will cease to execute. As for result, each time the code in the while block is executed, result is updated by multiplying … Read more
“While” loop isn’t necessary. Your code works perfectly fine without while loop, if you write like this: scanf(“%d”, &t); for (i = 1; i <= t; i++) If you want multiple “t” inputs you should move your return statement after while brace: int t, i; float h, l, w; while (scanf(“%d”, &t)) { for (i … Read more
Here’s a tip. You’ll want to start at the users number and count down to 0. Like this: int finalNum = 0; int userNum; //This is where you need to get the user’s number…. while(userNum > 0) { finalNum += userNum; userNum–; } //Do whatever you need to finalNum…. EDIT: It appears you’ve posted pseudocode; … Read more
This should work for your purposes: name = input(“Please enter your name: “) name = name.lower() luckynumb = 0 firstnamenumb = 0 surnamenumb = 0 number = [1, 2, 3, 4, 5, 6, 7, 8, 9] row1 = [“a”, “b”, “c”, “d”, “e”, “f”, “g”, “h”, “i”] row2 = [“j”, “k”, “l”, “m”, “n”, “o”, … Read more
It would be better if you use a for instead of a while, but if it’s an exercise requirement than you can write: $int = 8; $i = 1; $sum = 0; while ($i <= $int) { $sum += $i; $i++; } A more advanced notation would be: $sum += $i++; (so you can remove … Read more
I can see a couple of errors in your code. First, as Pete Becker noted (and Lightness Races in Orbit refined), total isn’t initialized, so its value is indeterminate. It could even be an invalid representation of a floating point number (e.g. a NaN ) and it isn’t for sure a valid starting point for … Read more
Every widget constructor should never block the main message loop! The main message loop looks usually like this: int main(int argc, char *argv[]) { QApplication a(argc, argv); MainWindow w(nullptr); w.show(); int r = a.exec(); return r } In your case your MainWindow ctor never returns, so w.show() is never called and a.exec() (main messgae loop) … Read more
Just add one line of code after the cout statement as follow: if (mode == TOPIG) { cout << “TOPIG MODE” << endl; while (cin.good()){ getline(cin, lines); while (!looking) { spot = lines.find(” “); if (spot == -1){ looking = true; spot = lines.length( ); } line = lines.substr(0, spot); TP1stLetter(line); if (!looking) lines = … Read more
The problem in your current code is that you’re always trying to read an int so when receiving a non-integer input you can’t handle the error in the right way. Modify this to always read a String and convert it into an int: int numberOfPeople = 0; while (numberOfPeople <= 0) { try { System.out.print(“Enter … Read more
The while loop will continue as long as temp is true. In the nested if-statement we set temp as false which exits the loop. As for the playerPick: the variable is declared outside of the loop so It should be accessible anywhere within the function that is below the declaration (code is read top down … Read more
You cannot really use the setTimeout() function as you suggest… I guess this is roughly what you are looking for: echo <<< EOT <script type=”text/javascript”> setTimeout(function() { window.open(‘https://mywebsite/$link’, ‘_blank’); }, 1000); </script> EOT; Note: I just use the nowdoc notation since it is easier to read. Certainly it is possible to use a normal literal … Read more
The problem is that you return the inputed string at the end of the loop no matter the validation done. You can use a boolean to check this validity. Also you don’t need to parse all the string and you can break the inner loop on the first invalid char. I renamed the string as … Read more
When you do cin >> divers; the end-of-line character is not removed from the input, just the number leading up to it. Then, the next time you ask for a line with std::getline() it returns just the end-of-line character that is already there and does not wait for your new input. So when you do … Read more
Wrap the result in another table. echo “<table>”; $count = 0; $num_columns = 2; // or 3 while ($rc = mysql_fetch_array($results_course)) { if ($count++ % $num_columns == 0) { echo “<tr>”; } echo “<td>”; // previous table code here echo “</td>”; if ($count % $num_columns == 0) { echo “</tr>”; } } if ($count % … Read more