string snack2[][]=new string[Math.Round(snacks.Length/2)][2]; int cnt=0; for(int i=0;i<Math.Round(snacks.Length/2);i++) { for (int j=0;j<2;j++) { snacks2[i][j]=snack[cnt++]; } } 3 solved How To convert 1D array to 2D array in C# [closed]
In the remove function, you must update tailif the list becomes empty. int removeFirst() { List *nodeToRemove; int dataToRemove; if (head->next == NULL) { prtError(“Empty list!”); return -1; } else { nodeToRemove = head->next; dataToRemove = nodeToRemove->data; head->next = nodeToRemove->next; free(nodeToRemove); // Update tail if list is empty if (head->next == NULL) { tail = … Read more
You can try this, by simply using char as binary operator and the Increment and Decrement can be applied on chars as well as on integer. If you want to get input from the user you can use the scanf(“%c”) to read the chars and use them char i,j; for ( i = ‘G’; i … Read more
Even though it’s hard to say what you’re trying to achieve with the code, I must say that it’s an ordering error: you want to append the newly created items to the #icons element, which you declare at the very end. Move the last statement to the top of the function and everything will work … Read more
When we try to create a String Object JVM first checks the String constant pool,if string doesn’t exist in the pool then it will creates a new String Object isaq and reference maintained in the pool.The Variable S1 also refer the same String Object. String s1=”isaq”; Above statement create one String Object in String pool. … Read more
You can use the modulus operator (%), this will help make your code future proof, should the 20 need to change to 200 etc for(var i = 1; i <= 20; i++){ if(i % 5 == 0 && i % 3 == 0) { console.log(“FizzBuzz”); } else if (i % 3 == 0) { console.log(“Fizz”); … Read more
First of all, toString method comes from object class, don’t try to make multiple of those methods or anything like that, have everything in one method. As to your problem, you can make a variable String description for your class, and initialize it in constructor and then in toString just return that variable. You can … Read more
There needs to be a body to those empty if/elif statements Example (python 2 style): if computerChoice == 0: print ‘Something needs to be done here’ elif computerChoice == 1: print ‘Something needs to be done here too’ elif computerChoice == 2: print ‘And something needs to be done here, too’ Obviously replace those print … Read more
Found the issue. The profiler is not recording included libraries. In the magento/lib folder nothing is recorded even if you explicitly set it to profile. It seems that we need to use an all-encompassing php profiler solved What can delay “Time to First byte” by almost 2s?
If it was for i in range(1,10): if space_check(board, i): return False else: return True then after the first iteration in the for loop the function would return. This would not lead the the expected behaviour. Currently, you check every space, and not just the first one 9 solved Are booleans overwritten in python?
echo $form->data[‘Itemid’]; Or if you mean inside the foreach loop (because you’ve got other stuff to do there), then use this: foreach($form->data as $key => $value) { if( $key === ‘Itemid’ ) echo $form->data[‘Itemid’]; } 2 solved display single value of php array
I can’t explain this Let me explain it for you. You are looking for pivoting the comments values for the concern for each student. Unfortunately, MySQL has no pivot table operator. However, you can use the CASE expression to do this. Like so: SELECT student_name, MAX(CASE WHEN concern = ‘Academics’ THEN comments END) AS ‘Accademics’, … Read more
You can’t declare a variable within an object literal. Move pid outside: var pid = … $.ajax({ … }); solved Chrome SyntaxError: Unexpected Identifier [closed]
Change your method to private string NumbersToWords(int number) { string word = “”; if (number == 8) { word = “Eight”; } return word; } Previously you did not return a value, indicated by void. 3 solved C# convert a number into words
Try this: $(document).ready(function (){ $.getJSON(‘myurl.com/myFile.json’,function(result){ alert(result); }); }); It should be $(document).ready not $(document).read Remove the extra a after the alert. Remove the extra bracket after the url ‘myurl.com/myFile.json’) <– This one Please have a look at the jQuery API before starting something new. 2 solved Cannot access json with getJson function