The first problem: def display(self): if SearchString in per.fname: print per.fname, per.lname elif SearchString in per.lname: print per.fname, per.lname else: print “No String” if statements do not form a group of alternatives themselves. To indicate that use the elif keyword, so the else part will be executed if none of the conditions were met. In … Read more
Python understands 1.2e34, as a float, but you can cast it to an int. int(1.2e34). If you want to loop between 1 and n inclusive, you would normally use range(1, n+1). Thus, in Python 3: for i in range(1, int(1.2e34)+1): print(i) # or do whatever you want — As FHTMitchell pointed out, in Python 2, … Read more
Code: int[] input = new int[]{0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 8, 8, 10, 10, 2, 2, 2, 3, 3, 7, 7}; int current = input[0]; boolean found = false; for (int i = 1; i < input.length; i++) { if (current == … Read more
You need two loops – one nested within another. One to print the rows and another to print the numbers within the rows. The range of the first loop is pretty straight forward, but the range of the nested loop will depend on the parent loop. For Python 3.x: for i in range(1,11): for j … Read more
Here’s a minimal, but working example. You can pass any type, to any method… for as long as the method is expecting the type as incoming parameter. private void DoSomething(string withThisString) In that method declaration, I declared that … the method is private aka accessible to this class only the method returns void aka does … Read more
As you wanted to display 3 elements from the array each time and when the last element is displayed you want to repeat the process again you are suppose to use setInterval() to call the function after a specified time continuously and reset the index to 0 immediately after the last element is displayed. Else … Read more
Let’s assume your input file contains the following as in your example: 1 + 2 + 5 – 3 = The outer scanf reads with %d. This picks up 1 as the integer value 1 and puts it in result. Now the inner scanf runs with %c. It reads the space after 1 so none … Read more
You need to break up the rows and convert each value to an integer. At the moment you are looking for the presence of the string “3” which is why strings like “2;13” pass the test. Try something like this: list_6 = [“4;99”, “3;4;8;9;14;18”, “2;3;8;12;18”, “2;3;11;18”, “2;3;8;18”, “2;3;4;5;6;7;8;9;11;12;15;16;17;18”, “2;3;4;8;9;10;11;13;18”, “1;3;4;5;6;7;13;16;17”, “2;3;4;5;6;7;8;9;11;12;14;15;18”, “3;11;18”, “2;3;5;8;9;11;12;13;15;16;17;18”, “2;5;11;18”, “1;2;3;4;5;8;9;11;17;18”, … Read more
while(true) is an infinite loop, the only way of getting out of it is using break. Or changing while(true) to some condition that alternatively ends. In this code, the while(true) part makes no real sense to me, you should probably add something else to that part as well, e.g. some connection stuff etc. 6 solved … Read more
32 million records is a large amount of almost anything, however if you are receiving the information from a Database perhaps there is a way to break it up into to parallel chunks. You could devise a strategy to execute a series of queries and combine the results. Take a look at the Java Future … Read more
Although it’s not entirely clear what you’re trying to achieve I have tried to modify your code from your original post to make it more readable and function the way you want. Please see my comments in the code below, I tried to prefix all of my comments with “EDIT” so you could easily identify … Read more
As mentioned by πάντα ῥεῖ in the comments, you can’t create new variable names dynamically at runtime in C++. All variable names must be known at compile-time. What you need to do here is use array indices instead. For example, change somedata and Do to std::vectors. Something like this: std::vector<bool> Do(5); // contains 5 bools, … Read more
The condition i==j+1 will not be true for i==2. This can be fixed by a couple of changes to the inner loop: #include <stdio.h> int main(void) { for (int i=2; i<100; i++) { for (int j=2; j<=i; j++) // Changed upper bound { if (i == j) // Changed condition and reversed order of if:s … Read more
public class Anagram { public static void main(String[] args) { String text = “!Hello123 “; char[] chars = text.toCharArray(); int left = 0; int right = text.length() – 1; while (left < right) { boolean isLeftLetter = Character.isLetter(chars[left]); boolean isRightLetter = Character.isLetter(chars[right]); if (isLeftLetter && isRightLetter) { swap(chars, left, right); left++; right–; } else { … Read more
You need to add a line as an element to an array. So, you need not array(), but array_push(). int array_push ( array &$array , mixed $value1 [, mixed $… ] ) or in your case, array_push($lineGroup, $line); also, you can use $lineGroup[] = $line; solved My loop produces the same vaues each time – … Read more