I was doing a JS quiz and stumbled uppon this question.
What does this function return?
function Batman()
{
return Array(4).join("lol" - 2) + " Batman!";
}
This actually happens to return NaNNaNNaN Batman! which i found rather funny.
But why does it return exactly this? i mean it stores NaN 3 times and suddenly it skips that and puts batman in the final index. While afaik the exact same thing happens for each array index.
I am not getting what you want to do.
Your function works like bellow.
Array(4) //retruns [undefined × 4]
You are saying
[undefined × 4].join("lol"-2)
Here because "lol"-2 returns NaN it will return "NaNNaNNaN"
And after this you are appendig the result to " Batman!" so the last result will be
"NaNNaNNaN Batman!"
The code tries to subtract 2 from "lol". But because "lol" isn't a number, the result isn't a number either; that gets represented internally by a value known as "Not a Number", or "NaN" for short.
You've got several of those in an array. When you try to join them all together as strings, and add on " Batman!" to the end, the NaN gets converted to its most sensible string representation, which is just NaN.
So you end up with several copies of "NaN", followed by "Batman!".
See Wikipedia's entry on NaN.
The expression Array(4).join("lol" - 2) returns a string consisting of 3 NaN. The + operator concatenates " Batman!" at the end of the previous string (it does not append any item to any array, since join() returns a string). So, the function returns that funny sentence.
Related
Recently I came across an interesting website that illustrates a Javascript Obfuscator: http://bl.ocks.org/jasonsperske/5400283
For example, (([]===[])+/-/)[1] gives a and (1+{})[(1<<1)+1] gives b.
I have tried hard to understand the evaluation sequence of these obfuscated result but was in vain.
Taking (1+{})[(1<<1)+1] as an example, I understand that << is the bitwise shift operator and will return 2, so the expression becomes (1+{})[3]. But then I cannot understand what does it mean by 1+{} and [3].
Google isn't really helpful to this problem as search engines don't like the brackets or slashes very much, so in case there are duplicate questions I'm sorry about that.
It's just obfuscation tricks.
for example :
[]===[] ===> false
and
([]===[])+/-/ ===> "false/-/" ( You could test it in the console by yourself)
So what is (([]===[])+/-/)[1] ? ( second char)
That's right :'a'
You may want to look at this also :
You could go step by step:
(([]===[]))
is simply false. Converted into a string "false/-/"and indexed by [1] gives you the a of the string "false".
The same goes for (1+{}) which results in the string "1[object Object]".
And 1<<1+1 is another way of writing 3 so this results in "1[object Object]"[3], which is simply b.
1+{}'s result is a string "1[object Object]", (1+{})[3] is to get the char of index 3 which is b.
The first example:
[]===[]
Comparing two different objects with ===, so the result is false, whose toString result is "false".
/-/ is a regex object, whose toString result is "/-/"
When you do false + /-/, which will concat using the result of .toString(), so the result will be "false/-/", and the second char is a.
The 'Wat' talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.
I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.
The behaviours specific to JavaScript (as I don't know Ruby) are listed below.
I found in the JSFiddle that some of my results didn't correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.
Empty Array + Empty Array
[] + []
result:
<Empty String>
I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video's result.
Empty Array + Object
[] + {}
result:
[Object]
This matches the video's result. What's going on here? Why is this an object. What does the + operator do?
Object + Empty Array
{} + []
result:
[Object]
This doesn't match the video. The video suggests that the result is 0, whereas I get [Object].
Object + Object
{} + {}
result:
[Object][Object]
This doesn't match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.
Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
Doing wat + 1 results in wat1wat1wat1wat1...
I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.
Here's a list of explanations for the results you're seeing (and supposed to be seeing). The references I'm using are from the ECMA-262 standard.
[] + []
When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.
[] + {}
Similar to [] + [], both operands are converted to primitives first. For "Object objects" (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).
{} + []
The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you're not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.
{} + {}
Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Array(16).join("wat" - 1)
As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that
If either operand is NaN, the result is NaN.
So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):
If m is NaN, return the String "NaN".
Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you're seeing.
When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").
As to why you're seeing different results for the {} + [] case: When using it as a function argument, you're forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it's instead parsed as an empty object literal.
This is more of a comment than an answer, but for some reason I can't comment on your question. I wanted to correct your JSFiddle code. However, I posted this on Hacker News and someone suggested that I repost it here.
The problem in the JSFiddle code is that ({}) (opening braces inside of parentheses) is not the same as {} (opening braces as the start of a line of code). So when you type out({} + []) you are forcing the {} to be something which it is not when you type {} + []. This is part of the overall 'wat'-ness of Javascript.
The basic idea was simple JavaScript wanted to allow both of these forms:
if (u)
v;
if (x) {
y;
z;
}
To do so, two interpretations were made of the opening brace: 1. it is not required and 2. it can appear anywhere.
This was a wrong move. Real code doesn't have an opening brace appearing in the middle of nowhere, and real code also tends to be more fragile when it uses the first form rather than the second. (About once every other month at my last job, I'd get called to a coworker's desk when their modifications to my code weren't working, and the problem was that they'd added a line to the "if" without adding curly braces. I eventually just adopted the habit that the curly braces are always required, even when you're only writing one line.)
Fortunately in many cases eval() will replicate the full wat-ness of JavaScript. The JSFiddle code should read:
function out(code) {
function format(x) {
return typeof x === "string" ?
JSON.stringify(x) : x;
}
document.writeln('>>> ' + code);
document.writeln(format(eval(code)));
}
document.writeln("<pre>");
out('[] + []');
out('[] + {}');
out('{} + []');
out('{} + {}');
out('Array(16).join("wat" + 1)');
out('Array(16).join("wat - 1")');
out('Array(16).join("wat" - 1) + " Batman!"');
document.writeln("</pre>");
[Also that is the first time I have written document.writeln in many many many years, and I feel a little dirty writing anything involving both document.writeln() and eval().]
I second #Ventero’s solution. If you want to, you can go into more detail as to how + converts its operands.
First step (§9.1): convert both operands to primitives (primitive values are undefined, null, booleans, numbers, strings; all other values are objects, including arrays and functions). If an operand is already primitive, you are done. If not, it is an object obj and the following steps are performed:
Call obj.valueOf(). If it returns a primitive, you are done. Direct instances of Object and arrays return themselves, so you are not done yet.
Call obj.toString(). If it returns a primitive, you are done. {} and [] both return a string, so you are done.
Otherwise, throw a TypeError.
For dates, step 1 and 2 are swapped. You can observe the conversion behavior as follows:
var obj = {
valueOf: function () {
console.log("valueOf");
return {}; // not a primitive
},
toString: function () {
console.log("toString");
return {}; // not a primitive
}
}
Interaction (Number() first converts to primitive then to number):
> Number(obj)
valueOf
toString
TypeError: Cannot convert object to primitive value
Second step (§11.6.1): If one of the operands is a string, the other operand is also converted to string and the result is produced by concatenating two strings. Otherwise, both operands are converted to numbers and the result is produced by adding them.
More detailed explanation of the conversion process: “What is {} + {} in JavaScript?”
We may refer to the specification and that's great and most accurate, but most of the cases can also be explained in a more comprehensible way with the following statements:
+ and - operators work only with primitive values. More specifically +(addition) works with either strings or numbers, and +(unary) and -(subtraction and unary) works only with numbers.
All native functions or operators that expect primitive value as argument, will first convert that argument to desired primitive type. It is done with valueOf or toString, which are available on any object. That's the reason why such functions or operators don't throw errors when invoked on objects.
So we may say that:
[] + [] is same as String([]) + String([]) which is same as '' + ''. I mentioned above that +(addition) is also valid for numbers, but there is no valid number representation of an array in JavaScript, so addition of strings is used instead.
[] + {} is same as String([]) + String({}) which is same as '' + '[object Object]'
{} + []. This one deserves more explanation (see Ventero answer). In that case, curly braces are treated not as an object but as an empty block, so it turns out to be same as +[]. Unary + works only with numbers, so the implementation tries to get a number out of []. First it tries valueOf which in the case of arrays returns the same object, so then it tries the last resort: conversion of a toString result to a number. We may write it as +Number(String([])) which is same as +Number('') which is same as +0.
Array(16).join("wat" - 1) subtraction - works only with numbers, so it's the same as: Array(16).join(Number("wat") - 1), as "wat" can't be converted to a valid number. We receive NaN, and any arithmetic operation on NaN results with NaN, so we have: Array(16).join(NaN).
To buttress what has been shared earlier.
The underlying cause of this behaviour is partly due to the weakly-typed nature of JavaScript. For example, the expression 1 + “2” is ambiguous since there are two possible interpretations based on the operand types (int, string) and (int int):
User intends to concatenate two strings, result: “12”
User intends to add two numbers, result: 3
Thus with varying input types,the output possibilities increase.
The addition algorithm
Coerce operands to primitive values
The JavaScript primitives are string, number, null, undefined and boolean (Symbol is coming soon in ES6). Any other value is an object (e.g. arrays, functions and objects). The coercion process for converting objects into primitive values is described thus:
If a primitive value is returned when object.valueOf() is invoked, then return this value, otherwise continue
If a primitive value is returned when object.toString() is invoked, then return this value, otherwise continue
Throw a TypeError
Note: For date values, the order is to invoke toString before valueOf.
If any operand value is a string, then do a string concatenation
Otherwise, convert both operands to their numeric value and then add these values
Knowing the various coercion values of types in JavaScript does help to make the confusing outputs clearer. See the coercion table below
+-----------------+-------------------+---------------+
| Primitive Value | String value | Numeric value |
+-----------------+-------------------+---------------+
| null | “null” | 0 |
| undefined | “undefined” | NaN |
| true | “true” | 1 |
| false | “false” | 0 |
| 123 | “123” | 123 |
| [] | “” | 0 |
| {} | “[object Object]” | NaN |
+-----------------+-------------------+---------------+
It is also good to know that JavaScript's + operator is left-associative as this determines what the output will be cases involving more than one + operation.
Leveraging the
Thus 1 + "2" will give "12" because any addition involving a string will always default to string concatenation.
You can read more examples in this blog post (disclaimer I wrote it).
This question already has answers here:
Why does ++[[]][+[]]+[+[]] return the string "10"?
(10 answers)
Closed 6 years ago.
I have recently seen an expression from a source, which looks something like below -
++[[]][+[]]+[+[]]
Entering this into the Chrome (Windows 7, Version 27.0.1453.94 m) console shows a result of "10".
Can someone explain what's happening here?
JSFiddle.
JavaScript is fairly flexible about converting between data types. The first thing to notice is that +[] evaluates to 0.* That lets us rewrite the expression as:
++[[]][0] + [0]
The next thing to notice is that ++[[]][0] is the preincrement operator applied to the first element of [[]]. Normally you can't apply ++ to an array, but JavaScript kindly converts the first element to 0, so the result is that ++[[]][0] evaluates to 1 (the first element of [[]] having now been incremented). It is kind of like this:
var a = [[]];
var b = ++a[0];
// now a will be [1] and b will be 1
That leaves us with:
1 + [0]
JavaScript now converts the int and the array to strings (since [0] is not a numeric value) and concatenates them together. Done!
* My understanding of how +[] becomes 0 is that it is a two-step process: first, [] is converted to a string primitive, which is the empty string. The empty string then converts to a number, which is zero. Via the same route, [1] evaluates to '1' and then to 1, [2] evaluates to 2, etc. However, [1, 2] evaluates to '1,2' which evaluates to NaN. (The last because the decimal point separator is ., not ,. I don't know what would happen if my locale were different.)
This expression stringifies valid Javascript constructs that yelds NaN, numbers, boolean undefined etc.
e.g.
+[] -> 0 //The unary plus operator is applied to the result of toString applied to an empty array (which is an empty string)
!+[] -> true
You can have a look also at this question,
and at the no alnum cheat sheets.
+[] is a number conversion from array to number which is 0.
and +[0] is also 0.
So the final result can be deduced to (++0) + [0] which is 1+[0].
And for a number adding an array. They are converted to string so the result is actually '10'.
You can log typeof(++[[]][+[]]+[+[]]) to verify.
What is the difference between parseInt(string) and Number(string) in JavaScript has been asked previously.
But the answers basically focused on the radix and the ability of parseInt to take a string like "123htg" and turn it into 123.
What I am asking here is if there is any big difference between the returns of Number(...) and parseFloat(...) when you pass it an actual number string with no radix at all.
The internal workings are not that different, as #James Allardic already answered. There is a difference though. Using parseFloat, a (trimmed) string starting with one or more numeric characters followed by alphanumeric characters can convert to a Number, with Number that will not succeed. As in:
parseFloat('3.23abc'); //=> 3.23
Number('3.23abc'); //=> NaN
In both conversions, the input string is trimmed, by the way:
parseFloat(' 3.23abc '); //=> 3.23
Number(' 3.23 '); //=> 3.23
No. Both will result in the internal ToNumber(string) function being called.
From ES5 section 15.7.1 (The Number Constructor Called as a Function):
When Number is called as a function rather than as a constructor, it performs a type conversion...
Returns a Number value (not a Number object) computed by ToNumber(value) if value was supplied, else returns +0.
From ES5 section 15.1.2.3 (parseFloat (string)):
...
If neither trimmedString nor any prefix of trimmedString satisfies the syntax of a StrDecimalLiteral (see 9.3.1)
...
And 9.3.1 is the section titled "ToNumber Applied to the String Type", which is what the first quote is referring to when it says ToNumber(value).
Update (see comments)
By calling the Number constructor with the new operator, you will get an instance of the Number object, rather than a numeric literal. For example:
typeof new Number(10); //object
typeof Number(10); //number
This is defined in section 15.7.2 (The Number Constructor):
When Number is called as part of a new expression it is a constructor: it initialises the newly created object.
Not a whole lot of difference, as long as you're sure there's nothing but digits in your string. If there are, Number will return NaN. Another problem that you might get using the Number constructor is that co-workers might think you forgot the new keyword, and add it later on, causing strict comparisons to fail new Number(123) === 123 --> false whereas Number(123) === 123 --> true.
In general, I prefer to leave the Number constructor for what it is, and just use the shortest syntax there is to cast to an int/float: +numString, or use parse*.
When not using new to create a wrapper object for a numerical value, Number is relegated to simply doing type conversion from string to number.
'parseFloat' on the other hand, as you mentioned, can parse a floating point number from any string that starts with a digit, a decimal, or +/-
So, if you're only working with strings that contain only numerical values, Number(x) and parseFloat(x) will result in the same values
Please excuse me posting yet another answer, but I just got here via a Google search and did not find all of the details that I wanted. Running the following code in Node.js:
var vals = ["1", "1.1", "0", "1.1abc", "", " ", null];
for(var i = 0; i < vals.length; i++){
var ifTest = false;
if(vals[i])
{
ifTest = true;
}
console.log("val=" + vals[i] + ", Number()=" + Number(vals[i])+ ", parseFloat()=" + parseFloat(vals[i]) + ", if()=" + ifTest);
}
gives the following output:
val=1, Number()=1, parseFloat()=1, if()=true
val=1.1, Number()=1.1, parseFloat()=1.1, if()=true
val=0, Number()=0, parseFloat()=0, if()=true
val=1.1abc, Number()=NaN, parseFloat()=1.1, if()=true
val=, Number()=0, parseFloat()=NaN, if()=false
val= , Number()=0, parseFloat()=NaN, if()=true
val=null, Number()=0, parseFloat()=NaN, if()=false
Some noteworthy takeaways:
If protecting with an if(val) before trying to convert to number, then parseFloat() will return a number except in the whitespace case.
Number returns a number in all cases except for non-numeric characters aside from white-space.
Please feel free to add any test cases that I may be missing.
when I run the javascript code below, I get the variable original as ending as
"1059823647undefinedundefinedundefinedundefinedundefinedundefinedundefinedundefinedundefinedundefined0"
why is this happening and how can i fix it?
original="012345678901234567890";
document.write("<textarea>");
document.write(original);
document.write("</textarea>"+"<br>");
/* scramble */
var scramble='1059823647';
scramble=scramble.split('');
var scrambled=new Array();
original=original.split('');
for(i=0;i<original.length;i++){
if(Math.round(Math.round(Math.floor(i/10)*10)+10)>=original.length){
scrambled[i]=original[i];
}else{
scrambled[i]=original[Math.round(Math.round(Math.floor(i/10)*10)+scramble[i%10])];
}
}
original='';
for(i=0;i<scrambled.length;i++){
original+=scrambled[i];
}
document.write("<textarea>");
document.write(original);
document.write("</textarea>"+"<br>");
undefined is being printed because your equation:
Math.round(Math.round(Math.floor(i/10)*10)+scramble[i%10])
is returning a number outside of the range of your array "original"
eg when i = 10, your equation returns 101.
I'm not entirely sure but i think what you mean to do is this:
(Math.floor(i/10)*10) + Number(scramble[i%10])
You're working with strings. But treating them like numbers. JavaScript will convert a string representation of a number to an actual number, but only when it needs to... And the + operator doesn't require such a conversion, as it acts as the concatenation operator for strings.
Therefore, this expression:
Math.round(Math.floor(i/10)*10)+scramble[i%10]
...is converting the first operand into a string and appending an element from the scramble array. You don't notice this for the first ten iterations, since when i<10 the first expression evaluates to 0... But after that, you're suddenly prefixing each scramble element with "10", and trying to access original indexes >= 100... of which there are none defined.
Solution:
Convert your strings to numbers before using them.
Math.round(Math.floor(i/10)*10)+ Number(scramble[i%10])