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The for loop isn't entered even if the initial requirement is true


The for loop isn't entered even if the initial requirement is true

By : Oscar Martinez
Date : October 24 2020, 06:10 PM
With these it helps I have the following function with a for loop inside it. The code is run on an Arduino and the Serial.print function shows that the function is entered correctly with the correct input value. But the for loop isn't entered. Does anyone have an idea why? , A for loop is usually constructed like this:
code :
for(init counter; condition; increase counter)
for (antalsteg = 0; antalsteg < steps; antalsteg++)


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I am trying to set a requirement to only enter an integer and if anything else ins entered show error message

I am trying to set a requirement to only enter an integer and if anything else ins entered show error message


By : user5859155
Date : March 29 2020, 07:55 AM
should help you out Use the fact that Integer.parseInt will throw a NumberFormatException if the inputted number isn't a real number. Catch that exception and then notify the user of the error.
code :
int partyIn;
try
{
   partyIn = Integer.parseInt(ans);
}
catch (NumberFormatException e)
{
   JOptionPane.showMessageDialog(null, "What you entered was not a number: " + ans);
}
ASP.NET Initial Load in Production - IPrinciple Isnt Set

ASP.NET Initial Load in Production - IPrinciple Isnt Set


By : user3063158
Date : March 29 2020, 07:55 AM
seems to work fine After having done a bit of experimentation it turned out that the problem was only manifesting in Google Chrome. All other browsers were performing as expected.
The solution was actually an oversight on my part. It seems Google Chrome has stricter rules around how cookies are dealt with in relation to domains.
Java line prints itself before its while(true) loop is entered

Java line prints itself before its while(true) loop is entered


By : sfritz
Date : March 29 2020, 07:55 AM
this one helps. This might be one of the reason that is happening with your code . You have already taken a UserInput from the keyboard object of the scanner class that's why it's giving the else response. This particularly happens when you take other than String input from that object
code :
public class Test {

    public static void main(String[] args) {
        Scanner keyboard= new Scanner(System.in);
        int n=keyboard.nextInt();

        String userResponse;
        while(true) {
            userResponse = keyboard.nextLine();
            if(userResponse.length() == 1 && userResponse.charAt(0) == 'y') {
                System.out.println("Great! Let's get started.");
                break;
            }
            else if(userResponse.length() == 1 && userResponse.charAt(0) == 'n') {
                System.out.println("Come back next time " + "" + ".");
                System.exit(0);
            }
            else {
                System.out.println("Invalid response.");
            }
        }
    }

} 
5
Invalid response.
public class Test {

    public static void main(String[] args) {
        Scanner keyboard= new Scanner(System.in);
        String n=keyboard.nextLine();
        String userResponse;
        while(true) {
            userResponse = keyboard.nextLine();
            if(userResponse.length() == 1 && userResponse.charAt(0) == 'y') {
                System.out.println("Great! Let's get started.");
                break;
            }
            else if(userResponse.length() == 1 && userResponse.charAt(0) == 'n') {
                System.out.println("Come back next time " + "" + ".");
                System.exit(0);
            }
            else {
                System.out.println("Invalid response.");
            }
        }
    }

}
j
y
Great! Let's get started.
public class Test {

    public static void main(String[] args) {
        Scanner keyboard= new Scanner(System.in);

        String userResponse;
        while(true) {
            userResponse = keyboard.nextLine();
            if(userResponse.length() == 1 && userResponse.charAt(0) == 'y') {
                System.out.println("Great! Let's get started.");
                break;
            }
            else if(userResponse.length() == 1 && userResponse.charAt(0) == 'n') {
                System.out.println("Come back next time " + "" + ".");
                System.exit(0);
            }
            else {
                System.out.println("Invalid response.");
            }
        }
    }

}
y
Great! Let's get started.
Spark GraphX : requirement failed: Invalid initial capacity

Spark GraphX : requirement failed: Invalid initial capacity


By : Fillipe Feitosa
Date : March 29 2020, 07:55 AM
will help you This appears to be a bug within Spark 2.0 (so far, have tested this against 2.0, 2.0.1, and 2.0.2). Jira [SPARK-18200]: GraphX Invalid initial capacity when running triangleCount has been created to address this.
Your code should work okay with Spark 1.6 as noted in this linked notebook.
while loop runs even if condition isnt true

while loop runs even if condition isnt true


By : John James
Date : March 29 2020, 07:55 AM
it should still fix some issue i am trying to make a calculator using c++, im trying to implement error handling, so if the user enters a non arithmetic operator, it will tell the user to please enter an operator, using a while loop. the problem is, even when the user enters an operator on the first run through, the while loop still executes. , You have a logic error:
code :
if ((op != "+") || (op != "-") || (op != "/") || (op != "*"))
if ((op != "+") && (op != "-") && (op != "/") && (op != "*"))
bool is_valid = (op == "+") || (op == "-") || (op == "/") || (op == "*");

if (!is_valid)
    check = 0;
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