One advice, if you don’t learn for yourself, then nobody can teach you. So, please try researching a bit on the internet, or reading a good book before you post a question. Assuming this is your first mistake, and you wont repeat this without working a little on your own, i’ll provide the code and … Read more
Assume the string follows the format <int,string>,…. Please find the pseudo-code below: Loop through the string `str` and { smaller_sign_pos = str.find(‘<‘, prev_pos) entry_comma_pos = str.find(‘,’, smaller_sign_pos+1) greater_sign_pos = str.find(‘>’, entry_comma_pos+1) if (all pos values are not `npos`) { int_value = atoi(str.substr(smaller_sign_pos+1, entry_comma_pos-smaller_sign_pos-1)) str_value = str.substr(entry_comma_pos+1, greater_sign_pos-entry_comma_pos-1) prev_pos = greater_sign_pos+1 append int_value to int array … Read more
The code for negatives is incorrect. You cannot just blindly negate the result when the base m is negative. , but . Also, you’re not printing anything if m is zero! And ints are signed by default so signed int is noise. floats are signed too; but here you could as well use a double … Read more
Here is a sample program. Enhance as required as suggested in the previous update. #include <stdio.h> #include <stdlib.h> int main () { int number, even, odd; int *evenarr, *oddarr; int evencount = 0, oddcount = 0; int i; char name; evenarr = (int *) malloc(100 * sizeof(int)); oddarr = (int *) malloc(100 * sizeof(int)); printf … Read more
If you want an int as result, multiply with the base: int a = 5; int b = 5; int c = 10 * a + b; // 55 If you want the result to be a std::string, use std::to_string (since C++11): int a = 5; int b = 5; std::string c = std::to_string(a) + … Read more
Try using, <?php $your_str = “12 , 13 , 9”; $array = explode(‘,’, $your_str); echo $array[0]; //12 echo $array[1]; //13 and so on… ?> explode works quite similar to split() which has been deprecated. Hope it helps! 1 solved PHP: Convert string to each integer variable
When you’re using Scanner.nextInt() the input is expected to be an integer. Inputting anything else, including a real number, will throw an InputMismatchException. To make sure invalid input doesn’t stop your program, use a try/catch to handle the exception: int num; try { num = sc.nextInt(); // Continue doing things with num } catch (InputMismatchException … Read more
Call stron the number and str.join the result of that: i = 4316 print(“+”.join(str(i))) 4+3+1+6 solved Python – How to show addition of integers digits
I think this link should give you an insight: Why Are Floating Point Numbers Inaccurate? It is understandable since the results can be slightly different due to precision if compared between how different languages deal with them. A quick example between using in python and just googling the answer: Python example Google example solved Why … Read more
var a = [“a”, “a”]; b = [“b”, “b”, “b”, “b”], c = [“c”, “c”, “c”], d = [“d”], container = [a, b, c, d]; ​container.sort(function (a, b) { return b.length – a.length; }); console.log(container); container will be sorted from longest (most elements) to shortest (least number of elements). Access it as: container[0] // longest … Read more
You can use the modulo operator to determine whether the word has an even or odd number of letters (% in java) if ( (word.length % 2) == 1 ) { //odd } else { //even } then just split the string in half and compare the reverse of the end with the front int … Read more
This is not binary, this is octal. In Ruby, any number starting with 0 will be treated as an octal number. You should check the Ruby number literals to learn more about this, here’s a quote: You can use a special prefix to write numbers in decimal, hexadecimal, octal or binary formats. For decimal numbers … Read more
mysqli_query does not return your number directly. It returns a mysqli_result as you can see here. To get the row you have parsed, you should fetch it first: $row = mysqli_fetch_assoc($result) A lot of information on using mysqli can be found here. 6 solved Object of class mysqli_result could not be converted to int – … Read more
For those with a maths background it is a surprising feature that System.out.println(“=” + (-3 % 4)); for example, returns -3 instead of 1 (which would be in the range [0,3]); Edit: in other words, the error message error message: Exception in thread “main” java.lang.ArrayIndexOutOfBoundsException: -8 is ultimately caused by (n % m) can return … Read more
A first step will be to convert the time in HH:MM:SS format (which is how your string is formatted) to that of seconds as per the following: String timerStop1 = String.format(“%02d”, hours) + “:” + String.format(“%02d”, minutes) + “:” + String.format(“%02d”, seconds); String[] timef=timerStop1.split(“:”); int hour=Integer.parseInt(timef[0]); int minute=Integer.parseInt(timef[1]); int second=Integer.parseInt(timef[2]); int temp; temp = second … Read more