octal

Please explain what, exactly, happens when the following sections of code are executed: int a='\15'; System.out.println(a); this prints out 13; int a='\25'; System.out.println(a); this prints out 21; int a='\100'; System.out.println(a); this prints out 64. Bohemian You have assigned a character literal, which is delimited by single quotes, eg 'a' (as distinct from a String literal, which is delimited by double quotes, eg "a" ) to an int variable. Java does an automatic widening cast from the 16-bit unsigned char to the 32-bit signed int . However, when a character literal is a backslash

I'm trying to build a php function and discovered some weird behavior and I can't even formulate a proper question, so if anyone can explain what is going on, I would appreciate it. I'm working with a set of numbers with leading zeros, and its important that they be maintained, but users almost never input leading zeros. So I use this: $x = 123; $n = 5; $x = str_pad((int)$x,$n,"0",STR_PAD_LEFT); echo $x; and, as desired, this gets me 00123. The weird stuff happens when I tested for a user inputting a zero before their number $x = 0123; $n = 5; $x = str_pad((int)$x,$n,"0",STR_PAD_LEFT); echo $x

I'm taking a numerical input as an argument and was just trying to account for leading zeroes. But it seems javascript converts the number into octal before I can do anything to the number. The only way to work around it so far is if I pass the number as a string initially but I was hoping there'd be another way to convert it after it is passed? So far tried (using 017 which alerted me to the octal behaviour): 017.toString(10) // 15 parseInt(017,10) // 15 017 + "" //15 new Number(017) //15 new Number('017') //17 parseInt('017', 10) // 17 So given function(numb) { if (typeof numb === number) {

What is the proper way to deal with leading zeros in Ruby? 0112.to_s => "74" 0112.to_i => 74 Why is it converting 0112 into 74 ? How can convert 0112 to a string "0112" ? I want to define a method that takes integer as a argument and returns it with its digits in descending order. But this does not seem to work for me when I have leading zeros: def descending_order(n) n.to_s.reverse.to_i end falsetru A numeric literal that starts with 0 is an octal representation, except the literals that start with 0x which represent hexadecimal numbers or 0b which represent binary numbers. 1 * 8**2 + 1 * 8*

I'm a beginner in python and I'm trying to use a octal number in my script, but when I try it, it returns me that error: >>> a = 010 SyntaxError: invalid token (<pyshell#0>, line 1) >>> 01 SyntaxError: invalid token (<pyshell#1>, line 1) There's something wrong with my code? I'm using Python3 (and reading a python 2.2 book) Try 0o10 , may be because of python 3, or pyshell itself. PEP says, octal literals must now be specified with a leading "0o" or "0O" instead of "0"; http://www.python.org/dev/peps/pep-3127/ 来源： https://stackoverflow.com/questions/1837874/invalid-token-when-using-octal

Why are Octal numeric literals not allowed in JavaScript strict mode ? What is the harm? "use strict"; var x = 010; //Uncaught SyntaxError: Octal literals are not allowed in strict mode. <h1>Check browser console for errors</h1> In case a developer needs to use Octals (which can mistakenly change a numbers meaning ), is there a workaround? The "why" part of the question is not really answerable. As for "how", off the top of my head... "use strict"; var x = parseInt('010', 8); document.write(x); Octal literals are not allowed because disallowing them discourages programmers from using leading