Reputation: 11764
What in the world is making the second parameter return true
?
WARNING: it will loop infinitely and might crash your browser
for(;;){
//...
}
I was totally expecting not to loop at all...
But it is running, and that makes it worse since it can only be running if something evaluated to true
, or am I missing something?
Upvotes: 5
Views: 232
Reputation: 26228
From for
MDN
for ([initialization]; [condition]; [final-expression])
statement
[Initialization]
and [final-expression]
are intuitively not required, reducing the construct to a simple while
loop. But the relevant part is:
condition
An expression to be evaluated before each loop iteration. If this expression evaluates to true, statement is executed. This conditional test is optional. If omitted, the condition always evaluates to true. If the expression evaluates to false, execution skips to the first expression following the for construct.
(emphasis mine)
This appears to be a totally arbitrary JS language design decision. If it were my language I would probably throw a syntax error.
On a playful side note, for(;;){}
is equivalent to while(true){}
, and happens to be 4 characters shorter. I wonder if minifiers leverage this!
Upvotes: 7
Reputation: 385870
There's nothing “making the second parameter return true
”. There is no second parameter. Section 12.6.3 of the ECMAScript Language Specification (ECMA-262 5th Edition) says this:
a. If the first Expression is present, then
i. Let testExprRef be the result of evaluating the first Expression.
ii. If GetValue(testExprRef) is false, return (
normal
, V,empty
).
The “first Expression” it's referring to is the optional expression between the semicolons of the for
statement. If it's present, the program performs steps (i) and (ii) quoted above. If the expression is not present, the program simply doesn't perform steps (i) and (ii) of the for
statement algorithm.
Upvotes: 1