I admit that I asked a question about why Closure Compiler does not shorten certain code which looks shortenable at first sight a few days ago already, but that reason is not applicable in this case and I'm not really sure why it isn't shortened here.
What code I have is:
var a = 0;
function b() {
return a++ >= 3;
}
Now there is pre-incrementing and post-incrementing. The difference is the return value - a++
returns a
a开发者_运维技巧nd then increments it, ++a
first increments a
and then returns it.
What this comes down to is that my code could be shortened to (ignoring whitespace removal):
var a = 0;
function b() {
return ++a > 3;
}
However, Closure Compiler does not seem to alter (or recognise) this.
My question therefore is: what side effects could ++a >
have when used instead of a++ >=
?
There is a particular edge-case for this construct (but not for 3).
It occurs because JavaScript stores numbers as IEEE-754 float-point 64-bit doubles and "only" has a guaranteed "exact" integer-representation up to 2^53 (although implementations may have lee-way to have a higher range, I do not know).
This is on Firefox 4:
a = 2e53
a++ >= 2e53 // true
a = 2e53
++a > 2e53 // false
Real question is what realized gain would such a very particular transformation have? :-0
Happy coding.
It’s safe to apply this size-optimisation if the right-operand (3
in your example) is a constant integer in the range [-252, 252]. In any other case (for example, if the right-operand is fractional or very large), it is not safe.
I would imagine that Closure does not implement this optimisation because:
- it requires a lot of checking to ensure that the optimisation is safe,
- it only applies in very specific circumstances that probably don’t come up very often, and
- It only saves a single character, which hardly seems worth the bother.
Why not check all of the edge conditions yourself?
function b(a) {
return a++ >= 3;
}
function b2(a) {
return ++a > 3;
}
console.log(b(2) === b2(2))
console.log(b(3) === b2(3))
console.log(b(4) === b2(4))
The output is true
in each case.
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