I have this line of code in PHP:
if($connect->query("INSERT INTO users (last_login) VALUES ('$now')") === TRUE) ...
and this line.
return true;
I've noticed that PHP booleans are sometimes uppercase and sometimes lowercase.
Why is this?
Should I use one for better performance?
Do they produce different results?
If so, what is the difference?
Are numeric booleans acceptable in PHP, because in JavaScript, booleans can be 0 and 1.
IE in JavaScript, you can do:
if(0) ... // false
if(1) ... // true
You can assign a truth value manually by stating that with the identity operator =.
$var = TRUE; // or true, is the same.
Anyway, boolean values are assigned automatically according to the nature of the variable you are considering. For example, a numeric value will be FALSE if it is 0, or TRUE in the other cases. A string will be FALSE just if empty.
PHP's rules on type-looseness and boolean conversions are fairly complex. From the docs (and Damien's comment): To specify a boolean literal, use the constants TRUE or FALSE. Both are case-insensitive.
Further down the page, under the "Converting to boolean" heading, all of the rules of numeric booleans, string booleans, etc. I'd recommend familiarizing yourself with them to avoid unintentional behavior. To answer your question directly: yes, numeric booleans are supported. And empty strings. And the string "0" will also count as false.
Related
I noticed something whilst debugging my code.
if(array.length=5){
console.log("it's 5 units long");
}
This not only makes the array size 5(assuming it performs the assignment every time) but it also performs the check in the if, resulting in a console output. Is this normal behaviour in Javascript and is this a valid shorthand for any real scenario?
The result of an assignment expression is the new value. Your code is equivalent to:
array.length = 5;
if (5) {
console.log("it's 5 units long");
}
And 5 is a truthy value, so the condition passes.
The following values are always falsy:
false
0 (zero)
"" (empty string)
null
undefined
NaN (a special Number value meaning Not-a-Number!)
All other values are truthy, including "0" (zero in quotes), "false" (false in quotes), empty functions, empty arrays, and empty objects.
I can contrive scenarios where this would be useful, but generally, no, it's not useful. At the very least, it's bad style, and shouldn't be used even if it can be.
I found these examples in another thread but I don't get it.
0 == '0' // true
0 to the left I converted to false(the only number that does that). The right is a non empty string which converts to true.
So how can
false == true --> true
What have I missed?
Here is an official answer to your question (the quoted parts, link at the bottom) and analysis:
Equality (==)
The equality operator converts the operands if they are not of the same type, then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the string operand is converted to a number if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
Syntax
x == y
Examples
3 == 3 // true
"3" == 3 // true
3 == '3' // true
This means, as I read it, that the first 3 (integer) is converted to string to satisfy the comparison, so it becomes '3' == '3', which is true, same as in your case with zeroes.
NOTE: I assume that the converion may vary in different JS engines, although they have to be unified under ECMAScript specifiction - http://www.ecma-international.org/ecma-262/5.1/#sec-11.9.3 (quoted #Derek朕會功夫). This assumption is made on a subjective and imperative opinion that not all browsers and JavaScript engines out there are ECMAScript compliant.
Identity / strict equality (===)
The identity operator returns true if the operands are strictly equal (see above) with no type conversion.
The Identity / strict equality (===) found on the same resource at the end of the answer will skip the automatic type conversion (as written above) and will perform type checking as well, to ensure that we have exact match, i.e. the expression above will fail on something like:
typeof(int) == typeof(string)
This is common operator in most languages with weak typing:
http://en.wikipedia.org/wiki/Strong_and_weak_typing
I would say that one should be certain what a function/method will return, if such function/method is about to return numbers (integers/floating point numbers) it should stick to that to the very end, otherwise opposite practices may cut your head off by many reasons and one lays in this question as well.
The above is valid for other languages with weak typing, too, like PHP for example.
Read more of this, refer second head (Equality operators):
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
When you use == JavaScript will try very hard to convert the two things you are trying to compare to the same type. In this case 0 is converted to '0' to do the comparison, which then results in true.
You can use ===, which will not do any type coercion and is best practice, to get the desired result.
Equality operator
The equality operator converts the operands if they are not of the same type, then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the string operand is converted to a number if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
Source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comparison_Operators
JavaScirpt Table Equality: http://dorey.github.io/JavaScript-Equality-Table/
I always thought that JavaScript's if statements did some kind of casting magic to their arguments, but I'm a little wary of what's actually going on behind the scenes.
I recently found a JavaScript comparison table and noticed that even though -1 == true evaluates to false, if(-1){...} will execute.
So within JavaScripts if statements, what happens to the expression? It seems reasonable to assume that it uses !!{expression} to cast it to an inverse boolean, then invert it again, but if that's the case, how does JS decide whether an object's inverse boolean representation is truthy or not?
JavaScript is wonky.
Yes, -1 == true results in false, but that's not what the if statement is doing. It's checking to see if the statement is 'truthy', or converts to true. In JavaScript, that's the equivalent of !!-1, which does result in true (all numbers other than zero are truthy).
Why?!?
The spec defines the double equals operator to do the following when presented with a number and a boolean:
If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).
ToNumber will convert the boolean true into the number 1, so you're comparing:
-1 == 1
which anyone can tell you is clearly false.
On the other hand, an if statement is calling ToBoolean, which considers any non-zero, non-NaN number to be true.
Any JavaScript developer really needs to look at the documentation -- for this case, located here: http://www.ecma-international.org/ecma-262/5.1/#sec-9.2
9.2 ToBoolean
The abstract operation ToBoolean converts its argument to a value of type Boolean according to Table 11:
Argument Type Result
Undefined false
Null false
Boolean The result equals the input argument (no conversion).
Number The result is false if the argument is +0, −0, or NaN; otherwise the result is true.
String The result is false if the argument is the empty String (its length is zero); otherwise the result is true.
Object true
(Sorry about the formatting, can't make a table here.)
From JavaScript The Definitive Guide
The following values convert to, and therefore work like, false:
undefined
null
0
-0
NaN
"" // the empty string
All other values, including all objects (and arrays) convert to, and work like, true. false, and the six values that convert to it, are sometimes called falsy values, and all other values are called truthy.
These things by themselves are falsy (or evaluate to false):
undefined
null
0
'' or ""
false
NaN
Everything else i truthy.
Truthy-ness or falsy-ness is used when evaluating a condition where the outcome is expected to be either truthy (true) or falsy (false).
In your example if(-1 == true), you are comparing apples and oranges. The compare is evaluated first (and resulted in false), and the results of that is used in your condition. The concept of truthyness/falsyness isn't applied to the operands the comparison.
When if state using with comparing variable different type js use .toString и .valueOf ( for more information check http://javascript.info/tutorial/object-conversion ) - just keep this in mind - it make so example much more easy to understand
console.log(true+true); //2
console.log(typeof(true+true)); //number
console.log(isNaN(true+true)); //false
Why is adding together 2 boolean types yielding a number? I kind of understand that if they didn't equal (1/0 (binary?)) it would be awkward to try to perform arithmetic on a boolean type, but I can't find the reasoning behind this logic.
It works like that because that's how it's specified to work.
EcmaScript standard specifies that unless either of the arguments is a string, the + operator is assumed to mean numeric addition and not string concatenation. Conversion to numeric values is explicitly mentioned:
Return the result of applying the addition operation to ToNumber( lprim) and ToNumber(rprim).
(where lprim and rprim are the primitive forms of the left-hand and the right-hand argument, respectively)
EcmaScript also specifies the To Number conversion for booleans clearly:
The result is 1 if the argument is true. The result is +0 if the argument is false.
Hence, true + true effectively means 1 + 1, or 2.
Javascript is loosely typed, and it automatically converts things into other things to fit the situation. That's why you can do var x without defining it as an int or bool
http://msdn.microsoft.com/en-us/library/6974wx4d(v=vs.94).aspx
Javascript is a dynamically typed language, because you don't have to specify what type something is when you start, like bool x or int i. When it sees an operation that can't really be done, it will convert the operands to whatever they need to be so that they can have that operation done on them. This is known as type coercion. You can't add booleans, so Javascript will cast the booleans to something that it can add, something like a string or a number. In this case, it makes sense to cast it to a number since 1 is often used to represent true and 0 for false. So Javascript will cast the true's to 1s, and add them together
Is it true that in if-statement JavaScript wrap the condition into a Boolean?
if(x) => if(Boolean(x))
Is it true that in comparison JavaScript wrap the compare elements into a Number?
a == b => Number(a) == Number(b)
Yes, and No.
For the first part, yes, that is essentially what the javascript does.
But for the latter, no. Not everything in JavaScript can be converted to a number. For example:
Number('abc') // => NaN
And Not-A-Numbers are not equal:
NaN == NaN // => false
So something like this:
Number('abc') == Number('abc') // => false!
But that's actually true with equality comparison.
'abc' == 'abc' // => true
As a side note, it's probably better to use === in JavaScript, which also checks the type of the values being compared:
0 == '0' // => true
0 === '0' // => false, because integer is not a string
More details about === can be read over here.
Yes, that's true, x is evaluated in a boolean context in this situation, so the equivalent of Boolean(x) is applied.
No, that's not true. It only looks that way because the coercitive equality operator == tries to convert a and b to the same type. Number() is only applied if either a or b is already a Number. For instance:
>>> 0x2A == 42
true // both 'a' and 'b' are numbers.
>>> "0x2A" == 42
true // 'a' is a string whose number coercion is equal to 'b'.
>>> "0x2A" == "42"
false // 'a' and 'b' are different strings.
Is it true that in if-statement JavaScript wrap the condition into a Boolean?
Usually yes.
Is it true that in comparison JavaScript wrap the compare elements into a Number?
Absolutely no.
Explanation
From JavaScript Language Specifications.
The if statement is defined at § 12.5 as:
if ( Expression ) Statement else Statement
It says that the Expression will be evaluated, converted with GetValue() and then tested after the ToBoolean() conversion.
Then the first assertion is true (but see later), the condition for the if statement is evaluated like is passed as parameter to the Boolean function. Please recall how JavaScript handles type conversion to boolean (§ 9.2):
undefined and null values are converted to false.
numbers are converted to false if ±0 or NaN otherwise they're converted to true.
strings are converted to false if empty otherwise always to true regardless their content.
objects are always converted to true.
Because of the call to GetValue() strictly speaking this assertion is not always true, take a look to § 8.7.1 where the standard describes how GetValue() works, here can happen some magic conversion before ToBoolean() is called.
The == operator is defined as in § 11.9.3.
As you can see it doesn't specify that operands must be (or will be treated as) numbers, the behavior of the operator is different and regulated by a series of rules based on the type of the operands. Then your second assertion is false. The case they're numbers (or one of them is a number) is just a special case in the algorithm, please note that at point 4 of the algorithm it says that if one of them is a number and the other one is a string then it'll be converted with ToNumber(), only in this case (with all the implications that this conversion has).
It's intuitive if you think that you can compare functions, strings or numbers, not every type can be converted to a numeric value.