JavaScript function 调用子表达式可以是尾调用吗？

Question

Consider the following return statement:

return f() || g();

The call f() obviously is not a tail call, because the function does not actually return if f() is falsy.

What about the g() part though, is that a tail call? Or do I have to rewrite it like this:

const temp = f();
if (temp) return temp; else return g();

Answer 1

Yes, but it doesn't help in practice.

According to the standard , g() is in tail position:

LogicalORExpression: LogicalORExpression || LogicalANDExpression

Return HasCallInTailPosition of LogicalANDExpression with argument call.

However, most browsers don't support tail call elimination, and the Chromium team isn't working on it, so you can't rely on tail call elimination in practice no matter how you write your tail calls.

Answer 2

Try it and see?

For some reason that didn't occur to me in my sleep-deprived state:D

function f() {
    return Math.random() > 1 || f();
}

f()

Node says RangeError: Maximum call stack size exceeded , Firefox says InternalError: too much recursion .