How to invoke a function on a String in javascript? - javascript

Maybe this is weird, but I have one more question for the same GA API helper function.
I need strings which will specify filter conditions, but I don't know how to use it properly.
Basically I want to generate and return a string such as "ga:medium==organic" from this function call medium.equals(organic) where var organic = "organic" and var medium = "ga:medium". I know how to work with the two strings to produce the output, and there are a lot more conditions to code for, but I just need to know how to invoke the function on the string medium and then how to use both the invoking object (string) and the arguments to return an output. Is what I'm asking possible?

You can add methods to the string object and then refer to this to access the string you're operating on.
String.prototype.equals = function(arg) {
return this + "==" + arg;
}
var medium = "ga:medium";
var organic = "organic";
var output = medium.equals(organic); // "ga:medium==organic"
Working demo: http://jsfiddle.net/jfriend00/hNLy6/

If your question is to compare two string logically and not as objects:
You can make use of the "===" operator which compares two operands for logical equality as
well as the type of the operands.
Example:
var x="5";
x==="5" // returns true
x===5 // returns false
Refer to this: http://www.w3schools.com/js/js_comparisons.asp

Related

Return primitive value of Javascript Number

I have a simple String extension to convert to numbers:
String.prototype.toInt = function(){ return parseInt(this); }
Because it's rather difficult without comments to know what kind of value you are getting in JS, I want to protect myself if I try to cast a Number to a Number that I thought was a string. I thought this would work:
Number.prototype.toInt = function(){ return this;}
But this returns the entire function. I see that there is a [[PrimitiveValue]] property in Number that ideally I could return, but I cannot figure out how to access it (such as this['[[PrimitiveValue]]'])
This should do the trick for you:
Number.prototype.toInt = function (){ return this.valueOf(); };
Though I must say, modifying the prototypes of native objects (Array, String, Number, etc) has many downsides. A functional approach is more flexible and fault tolerant, ie use parseInt(value) instead of value.toInt() would tolerate strings and numbers without issue.
Your method is returning the value as a boxed Number instance, so you just need to cast it to a primitive. For example, using the unary plus operator:
Number.prototype.toInt = function(){
return +this;
}

JavaScript chained concatenation function?

No jQuery please!
The Web says that the native String.concat() and join() functions of JS are to be avoided because of their poor performance, and a simple for() loop of += assignments should work a lot faster.
So I'm trying to create a function in pure JavaScript that will concatenate strings. This is somewhat how I envision it:
I want a main function concatenate() that will concatenate all passed arguments and additionally insert a variable string after each concatenated argument, except for the last one.
If the main function is called by itself and without the chained .using() function, then that variable string should be an empty one, which means no separators in the result.
I want a chained sub-function .using() that will tell the main concatenate() function what certain string other than the default '' empty string to add after each concatenated segment.
In theory, it should work like this:
concatenate('a','b','c'); /* result: 'abc' */
concatenate('a','b','c').using('-'); /* result: 'a-b-c' */
I want to avoid having two separate functions, like concatenate() and concatenateUsing(), because the concatenateUsing() variant would then have to utilize a special constant argument (like arguments[0] or arguments[arguments.length-1]) as the injected separator and that would be terribly untidy. Plus, I would always forget which one it was.
I also want to avoid having a superceding Concatenate object with two separate sub-methods, like Concatenate.strings() and Concatenate.using() or similar.
Here are some of my failed attempts so far...
Attempt #1:
function concatenate()
{
var result="";
if(this.separator===undefined){var separator=false;}
for(var i=0; i<arguments.length; i++)
{result += arguments[i] + ((separator && (i<arguments.length-1))?separator:'');}
this.using=function(x)
{
this.separator=x;
return this;
}
return result;
}
So what I'm trying to do is:
check if the separator variable is undefined, this means it wasn't set from a sub-method yet.
If it's undefined, declare it with the value false for later evaluation.
Run the concatenation, and if separator has another value than false then use it in each concatenation step - as long as it's not the last iteration.
Then return the result.
The sub-method .using(x) should somewhere along the way set the
value of the separator variable.
Naturally, this doesn't work.
Attempt #2:
var concatenate = function()
{
var result="";
var separator="";
for(var i=0; i<arguments.length; i++)
{result += arguments[i] + ((separator && (i<arguments.length-1))?separator:'');}
return result;
}
concatenate.prototype.using=function(x)
{
this.separator=x;
return this;
}
It also doesn't work, I assume that when this is returned from the using() sub-method, the var separator="" of the main concatenate() function just overwrites the value with "" again.
I tried doing this 4 or 5 different ways now, but I don't want to bore you with all the others as well.
Does anyone know a solution for this puzzle?
Thanks a lot in advance!
What you are trying to do is impossible.
You cannot chain something to a method call that returns a primitive, because primitives do not have (custom) methods1.
And you cannot make the first function return different things depending on whether something is chained or not, because it doesn't know about its call context and has to return the result before the method call is evaluated.
Your best bet is to return an object that can be stringified using a custom toString method, and also offers that using thing. It would be something along the lines of
function concatenate() {
return {
args: Array.from(arguments), // ES6 for simplicity
using: function(separator) {
return this.args.join(separator);
},
toString: function() {
return this.args.join("");
}
};
}
console.log(String(concatenate('a','b','c')); // result: 'abc'
// alternatively, use ""+… or explicitly call the ….toString() method
console.log(concatenate('a','b','c').using('-')); // result: 'a-b-c'
1: No, you don't want to know workarounds.

What is the difference between String and new String? [duplicate]

Taken from MDN
String literals (denoted by double or single quotes) and strings
returned from String calls in a non-constructor context (i.e., without
using the new keyword) are primitive strings. JavaScript automatically
converts primitives to String objects, so that it's possible to use
String object methods for primitive strings. In contexts where a
method is to be invoked on a primitive string or a property lookup
occurs, JavaScript will automatically wrap the string primitive and
call the method or perform the property lookup.
So, I thought (logically) operations (method calls) on string primitives should be slower than operations on string Objects because any string primitive is converted to string Object (extra work) before the method being applied on the string.
But in this test case, the result is opposite. The code block-1 runs faster than the code block-2, both code blocks are given below:
code block-1 :
var s = '0123456789';
for (var i = 0; i < s.length; i++) {
s.charAt(i);
}
code block-2 :
var s = new String('0123456789');
for (var i = 0; i < s.length; i++) {
s.charAt(i);
}
The results varies in browsers but the code block-1 is always faster. Can anyone please explain this, why the code block-1 is faster than code block-2.
JavaScript has two main type categories, primitives and objects.
var s = 'test';
var ss = new String('test');
The single quote/double quote patterns are identical in terms of functionality. That aside, the behaviour you are trying to name is called auto-boxing. So what actually happens is that a primitive is converted to its wrapper type when a method of the wrapper type is invoked. Put simple:
var s = 'test';
Is a primitive data type. It has no methods, it is nothing more than a pointer to a raw data memory reference, which explains the much faster random access speed.
So what happens when you do s.charAt(i) for instance?
Since s is not an instance of String, JavaScript will auto-box s, which has typeof string to its wrapper type, String, with typeof object or more precisely s.valueOf(s).prototype.toString.call = [object String].
The auto-boxing behaviour casts s back and forth to its wrapper type as needed, but the standard operations are incredibly fast since you are dealing with a simpler data type. However auto-boxing and Object.prototype.valueOf have different effects.
If you want to force the auto-boxing or to cast a primitive to its wrapper type, you can use Object.prototype.valueOf, but the behaviour is different. Based on a wide variety of test scenarios auto-boxing only applies the 'required' methods, without altering the primitive nature of the variable. Which is why you get better speed.
This is rather implementation-dependent, but I'll take a shot. I'll exemplify with V8 but I assume other engines use similar approaches.
A string primitive is parsed to a v8::String object. Hence, methods can be invoked directly on it as mentioned by jfriend00.
A String object, in the other hand, is parsed to a v8::StringObject which extends Object and, apart from being a full fledged object, serves as a wrapper for v8::String.
Now it is only logical, a call to new String('').method() has to unbox this v8::StringObject's v8::String before executing the method, hence it is slower.
In many other languages, primitive values do not have methods.
The way MDN puts it seems to be the simplest way to explain how primitives' auto-boxing works (as also mentioned in flav's answer), that is, how JavaScript's primitive-y values can invoke methods.
However, a smart engine will not convert a string primitive-y to String object every time you need to call a method. This is also informatively mentioned in the Annotated ES5 spec. with regard to resolving properties (and "methods"¹) of primitive values:
NOTE The object that may be created in step 1 is not accessible outside of the above method. An implementation might choose to avoid the actual creation of the object. [...]
At very low level, Strings are most often implemented as immutable scalar values. Example wrapper structure:
StringObject > String (> ...) > char[]
The more far you're from the primitive, the longer it will take to get to it. In practice, String primitives are much more frequent than StringObjects, hence it is not a surprise for engines to add methods to the String primitives' corresponding (interpreted) objects' Class instead of converting back and forth between String and StringObject as MDN's explanation suggests.
¹ In JavaScript, "method" is just a naming convention for a property which resolves to a value of type function.
In case of string literal we cannot assign properties
var x = "hello" ;
x.y = "world";
console.log(x.y); // this will print undefined
Whereas in case of String Object we can assign properties
var x = new String("hello");
x.y = "world";
console.log(x.y); // this will print world
String Literal:
String literals are immutable, which means, once they are created, their state can't be changed, which also makes them thread safe.
var a = 's';
var b = 's';
a==b result will be 'true' both string refer's same object.
String Object:
Here, two different objects are created, and they have different references:
var a = new String("s");
var b = new String("s");
a==b result will be false, because they have different references.
If you use new, you're explicitly stating that you want to create an instance of an Object. Therefore, new String is producing an Object wrapping the String primitive, which means any action on it involves an extra layer of work.
typeof new String(); // "object"
typeof ''; // "string"
As they are of different types, your JavaScript interpreter may also optimise them differently, as mentioned in comments.
When you declare:
var s = '0123456789';
you create a string primitive. That string primitive has methods that let you call methods on it without converting the primitive to a first class object. So your supposition that this would be slower because the string has to be converted to an object is not correct. It does not have to be converted to an object. The primitive itself can invoke the methods.
Converting it to an full-blown object (which allows you to add new properties to it) is an extra step and does not make the string oeprations faster (in fact your test shows that it makes them slower).
I can see that this question has been resolved long ago, there is another subtle distinction between string literals and string objects, as nobody seems to have touched on it, I thought I'd just write it for completeness.
Basically another distinction between the two is when using eval. eval('1 + 1') gives 2, whereas eval(new String('1 + 1')) gives '1 + 1', so if certain block of code can be executed both 'normally' or with eval, it could lead to weird results
The existence of an object has little to do with the actual behaviour of a String in ECMAScript/JavaScript engines as the root scope will simply contain function objects for this. So the charAt(int) function in case of a string literal will be searched and executed.
With a real object you add one more layer where the charAt(int) method also are searched on the object itself before the standard behaviour kicks in (same as above). Apparently there is a surprisingly large amount of work done in this case.
BTW I don't think that primitives are actually converted into Objects but the script engine will simply mark this variable as string type and therefore it can find all provided functions for it so it looks like you invoke an object. Don't forget this is a script runtime which works on different principles than an OO runtime.
The biggest difference between a string primitive and a string object is that objects must follow this rule for the == operator:
An expression comparing Objects is only true if the operands reference
the same Object.
So, whereas string primitives have a convenient == that compares the value, you're out of luck when it comes to making any other immutable object type (including a string object) behave like a value type.
"hello" == "hello"
-> true
new String("hello") == new String("hello") // beware!
-> false
(Others have noted that a string object is technically mutable because you can add properties to it. But it's not clear what that's useful for; the string value itself is not mutable.)
The code is optimized before running by the javascript engine.
In general, micro benchmarks can be misleading because compilers and interpreters rearrange, modify, remove and perform other tricks on parts of your code to make it run faster.
In other words, the written code tells what is the goal but the compiler and/or runtime will decide how to achieve that goal.
Block 1 is faster mainly because of:
var s = '0123456789'; is always faster than
var s = new String('0123456789');
because of the overhead of object creation.
The loop portion is not the one causing the slowdown because the chartAt() can be inlined by the interpreter.
Try removing the loop and rerun the test, you will see the speed ratio will be the same as if the loop were not removed. In other words, for these tests, the loop blocks at execution time have exactly the same bytecode/machine code.
For these types of micro benchmarks, looking at the bytecode or machine code wil provide a clearer picture.
we can define String in 3-ways
var a = "first way";
var b = String("second way");
var c = new String("third way");
// also we can create using
4. var d = a + '';
Check the type of the strings created using typeof operator
typeof a // "string"
typeof b // "string"
typeof c // "object"
when you compare a and b var
a==b ( // yes)
when you compare String object
var StringObj = new String("third way")
var StringObj2 = new String("third way")
StringObj == StringObj2 // no result will be false, because they have different references
In Javascript, primitive data types such is string is a non-composite building block. This means that they are just values, nothing more:
let a = "string value";
By default there is no built-in methods like toUpperCase, toLowerCase etc...
But, if you try to write:
console.log( a.toUpperCase() ); or console.log( a.toLowerCase() );
This will not throw any error, instead they will work as they should.
What happened ?
Well, when you try to access a property of a string a Javascript coerces string to an object by new String(a); known as wrapper object.
This process is linked to concept called function constructors in Javascript, where functions are used to create new objects.
When you type new String('String value'); here String is function constructor, which takes an argument and creates an empty object inside the function scope, this empty object is assigned to this and in this case, String supplies all those known built in functions we mentioned before. and as soon as operation is completed, for example do uppercase operation, wrapper object is discarded.
To prove that, let's do this:
let justString = 'Hello From String Value';
justString.addNewProperty = 'Added New Property';
console.log( justString );
Here output will be undefined. Why ?
In this case Javascript creates wrapper String object, sets new property addNewProperty and discards the wrapper object immediately. this is why you get undefined. Pseudo code would be look like this:
let justString = 'Hello From String Value';
let wrapperObject = new String( justString );
wrapperObject.addNewProperty = 'Added New Property'; //Do operation and discard

How to compare two regexps?

Why do
console.log(/a/ == /a/);
and
var regexp1 = /a/;
var regexp2 = /a/;
console.log(regexp1 == regexp2);
both return false?
Try this:
String(regexp1) === String(regexp2))
You are getting false because those two are different objects.
"Problem":
regex is an object- a reference type, so the comparsion is done by reference, and those are two different objects.
console.log(typeof /a/); // "object"
If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
MDN
Solution:
​var a = /a/;
var b = /a/;
console.log(​​​a.toString() === b.toString()); // true! yessss!
Live DEMO
Another "hack" to force the toString() on the regexes is:
console.log(a + "" === b + "");​
Just a guess - but doesn't JavaScript create a RegExp object for your regex, and therefore because you have created two different objects (even though they have the same "value") they're actually different?
For primitive data types like int, string, boolean javascript knows what to compare, but for objects like date or regex that operator only looks at the place in memory, because you define your regexes independently they have two different places in memory so they are not equal.

Replace a Regex capture group with uppercase in Javascript

I'd like to know how to replace a capture group with its uppercase in JavaScript. Here's a simplified version of what I've tried so far that's not working:
> a="foobar"
'foobar'
> a.replace( /(f)/, "$1".toUpperCase() )
'foobar'
> a.replace( /(f)/, String.prototype.toUpperCase.apply("$1") )
'foobar'
Would you explain what's wrong with this code?
You can pass a function to replace.
var r = a.replace(/(f)/, function(v) { return v.toUpperCase(); });
Explanation
a.replace( /(f)/, "$1".toUpperCase())
In this example you pass a string to the replace function. Since you are using the special replace syntax ($N grabs the Nth capture) you are simply giving the same value. The toUpperCase is actually deceiving because you are only making the replace string upper case (Which is somewhat pointless because the $ and one 1 characters have no upper case so the return value will still be "$1").
a.replace( /(f)/, String.prototype.toUpperCase.apply("$1"))
Believe it or not the semantics of this expression are exactly the same.
I know I'm late to the party but here is a shorter method that is more along the lines of your initial attempts.
a.replace('f', String.call.bind(a.toUpperCase));
So where did you go wrong and what is this new voodoo?
Problem 1
As stated before, you were attempting to pass the results of a called method as the second parameter of String.prototype.replace(), when instead you ought to be passing a reference to a function
Solution 1
That's easy enough to solve. Simply removing the parameters and parentheses will give us a reference rather than executing the function.
a.replace('f', String.prototype.toUpperCase.apply)
Problem 2
If you attempt to run the code now you will get an error stating that undefined is not a function and therefore cannot be called. This is because String.prototype.toUpperCase.apply is actually a reference to Function.prototype.apply() via JavaScript's prototypical inheritance. So what we are actually doing looks more like this
a.replace('f', Function.prototype.apply)
Which is obviously not what we have intended. How does it know to run Function.prototype.apply() on String.prototype.toUpperCase()?
Solution 2
Using Function.prototype.bind() we can create a copy of Function.prototype.call with its context specifically set to String.prototype.toUpperCase. We now have the following
a.replace('f', Function.prototype.apply.bind(String.prototype.toUpperCase))
Problem 3
The last issue is that String.prototype.replace() will pass several arguments to its replacement function. However, Function.prototype.apply() expects the second parameter to be an array but instead gets either a string or number (depending on if you use capture groups or not). This would cause an invalid argument list error.
Solution 3
Luckily, we can simply substitute in Function.prototype.call() (which accepts any number of arguments, none of which have type restrictions) for Function.prototype.apply(). We have now arrived at working code!
a.replace(/f/, Function.prototype.call.bind(String.prototype.toUpperCase))
Shedding bytes!
Nobody wants to type prototype a bunch of times. Instead we'll leverage the fact that we have objects that reference the same methods via inheritance. The String constructor, being a function, inherits from Function's prototype. This means that we can substitute in String.call for Function.prototype.call (actually we can use Date.call to save even more bytes but that's less semantic).
We can also leverage our variable 'a' since it's prototype includes a reference to String.prototype.toUpperCase we can swap that out with a.toUpperCase. It is the combination of the 3 solutions above and these byte saving measures that is how we get the code at the top of this post.
Why don't we just look up the definition?
If we write:
a.replace(/(f)/, x => x.toUpperCase())
we might as well just say:
a.replace('f','F')
Worse, I suspect nobody realises that their examples have been working only because they were capturing the whole regex with parentheses. If you look at the definition, the first parameter passed to the replacer function is actually the whole matched pattern and not the pattern you captured with parentheses:
function replacer(match, p1, p2, p3, offset, string)
If you want to use the arrow function notation:
a.replace(/xxx(yyy)zzz/, (match, p1) => p1.toUpperCase()
Old post but it worth to extend #ChaosPandion answer for other use cases with more restricted RegEx. E.g. ensure the (f) or capturing group surround with a specific format /z(f)oo/:
> a="foobazfoobar"
'foobazfoobar'
> a.replace(/z(f)oo/, function($0,$1) {return $0.replace($1, $1.toUpperCase());})
'foobazFoobar'
// Improve the RegEx so `(f)` will only get replaced when it begins with a dot or new line, etc.
I just want to highlight the two parameters of function makes finding a specific format and replacing a capturing group within the format possible.
SOLUTION
a.replace(/(f)/,(m,g)=>g.toUpperCase())
for replace all grup occurrences use /(f)/g regexp. The problem in your code: String.prototype.toUpperCase.apply("$1") and "$1".toUpperCase() gives "$1" (try in console by yourself) - so it not change anything and in fact you call twice a.replace( /(f)/, "$1") (which also change nothing).
let a= "foobar";
let b= a.replace(/(f)/,(m,g)=>g.toUpperCase());
let c= a.replace(/(o)/g,(m,g)=>g.toUpperCase());
console.log("/(f)/ ", b);
console.log("/(o)/g", c);
Given a dictionary (object, in this case, a Map) of property, values, and using .bind() as described at answers
const regex = /([A-z0-9]+)/;
const dictionary = new Map([["hello", 123]]);
let str = "hello";
str = str.replace(regex, dictionary.get.bind(dictionary));
console.log(str);
Using a JavaScript plain object and with a function defined to get return matched property value of the object, or original string if no match is found
const regex = /([A-z0-9]+)/;
const dictionary = {
"hello": 123,
[Symbol("dictionary")](prop) {
return this[prop] || prop
}
};
let str = "hello";
str = str.replace(regex, dictionary[Object.getOwnPropertySymbols(dictionary)[0]].bind(dictionary));
console.log(str);
In the case of string conversion from CamelCase to bash_case (ie: for filenames), use a callback with ternary operator.
The captured group selected with a regexp () in the first (left) replace arg is sent to the second (right) arg that is a callback function.
x and y give the captured string (don't know why 2 times!) and index (the third one) gives the index of the beginning of the captured group in the reference string.
Therefor a ternary operator can be used not to place _ at first occurence.
let str = 'MyStringName';
str = str.replace(/([^a-z0-9])/g, (x,y,index) => {
return index != 0 ? '_' + x.toLowerCase() : x.toLowerCase();
});
console.log(str);

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