digit

How do I remove the first digit of an integer? My input is an integer (for example i = 123456789). I then want to remove the first digit, so that i equals 23456789. try this n = n % (int) Math.pow(10, (int) Math.log10(n)); Here is one way to do it: Convert it to String Take the substring without the first "digit" Convert it to int Code: public static void main(String[] args) { int x = 123456789; String x_str = Integer.toString(x); int new_x = Integer.parseInt(x_str.substring(1)); System.out.println(new_x); } Output: 23456789 Note: This can be done in one line with int x = 123456789; int new_x

How can i get the last digit of an integer (or NSInteger) outputted to integer? example: int time = CFAbsoluteGetCurrent(); int lastDigit; Use modulo : int lastDigit = time % 10; 来源： https://stackoverflow.com/questions/4559654/c-objective-c-read-and-get-last-digit-of-integer

I'm trying to find the n th digit of an integer of an arbitrary length. I was going to convert the integer to a string and use the character at index n... char Digit = itoa(Number).at(n); ...But then I realized the itoa function isn't standard. Is there any other way to do this? (number/intPower(10, n))%10 just define the function intPower . You can also use the % operator and / for integer division in a loop. (Given integer n >= 0, n % 10 gives the units digit, and n / 10 chops off the units digit.) You can use ostringstream to convert to a text string, but a function along the lines of: char

Say I have a multi-digit integer in C. I want to break it up into single-digit integers. 123 would turn into 1 , 2 , and 3 . How can I do this, especially if I don't know how many digits the integer has? int value = 123; while (value > 0) { int digit = value % 10; // do something with digit value /= 10; } First, count the digits: unsigned int count(unsigned int i) { unsigned int ret=1; while (i/=10) ret++; return ret; } Then, you can store them in an array: unsigned int num=123; //for example unsigned int dig=count(num); char arr[dig]; while (dig--) { arr[dig]=num%10; num/=10; } As a hint,

for my selenium tests I need an value provider to get a 5-digit number in every case. The problem with javascript is that the api of Math.random only supports the generation of an 0. starting float. So it has to be between 10000 and 99999 . So it would be easy if it would only generates 0.10000 and higher, but it also generates 0.01000 . So this approach doesn't succeed: Math.floor(Math.random()*100000+1) Is it possible to generate a 5-digit number in every case (in an expression!) ? What about: Math.floor(Math.random()*90000) + 10000; Yes, you can create random numbers in any given range: var

As of C++14, thanks to n3781 (which in itself does not answer this question) we may write code like the following: const int x = 1'234; // one thousand two hundred and thirty four The aim is to improve on code like this: const int y = 100000000; and make it more readable. The underscore ( _ ) character was already taken in C++11 by user-defined literals, and the comma ( , ) has localisation problems — many European countries bafflingly † use this as the decimal separator — and conflicts with the comma operator, though I do wonder what real-world code could possibly have been broken by allowing

I wish to change an integer such as 23457689 to 689, 12457245 to 245 etc. I do not require the numbers to be rounded and do not wish to have to convert to String. Any ideas how this can be done in Python 2.7? Use the % operation: >>> x = 23457689 >>> x % 1000 689 % is the mod (i.e. modulo ) operation. To handle both positive and negative integers correctly: >>> x = -23457689 >>> print abs(x) % 1000 689 As a function where you can select the number of leading digits to keep: import math def extract_digits(integer, digits=3, keep_sign=False): sign = 1 if not keep_sign else int(math.copysign(1,