*   A new scheduling queue helps schedule the highest priority pending pod first. ([#55109](https://github.com/kubernetes/kubernetes/pull/55109),[ @bsalamat](https://github.com/bsalamat))

day!  Why, I haven't had a wink of sleep these three weeks!'

  `I'm very sorry you've been annoyed,' said Alice, who was
beginning to see its meaning.

  `And just as I'd taken the highest tree in the wood,' continued
the Pigeon, raising its voice to a shriek, `and just as I was
thinking I should be free of them at last, they must needs come
wriggling down from the sky!  Ugh, Serpent!'

- Added scoring for the prioritized list feature so nodes that best satisfy the highest-ranked subrequests were chosen. ([#134711](https://github.com/kubernetes/kubernetes/pull/134711), [@mortent](https://github.com/mortent)) [SIG Node, Scheduling and Testing]

n = len(m.nat.limbs). func rr(m *Modulus) *Nat { rr := NewNat().ExpandFor(m) n := uint(len(rr.limbs)) mLen := uint(m.BitLen()) logR := _W * n // We start by computing R = 2^(_W * n) mod m. We can get pretty close, to // 2^⌊log₂m⌋, by setting the highest bit we can without having to reduce. rr.limbs[n-1] = 1 << ((mLen - 1) % _W) // Then we double until we reach 2^(_W * n). for i := mLen - 1; i < logR; i++ { rr.Add(rr, m) } // Next we need to get from R to 2^(_W * n) R mod m (aka from one to R in //...

var bb = make([]byte, 1.0<<s)  // 1.0 has type int; len(bb) == 0
</pre>

<h4 id="Operator_precedence">Operator precedence</h4>
<p>
Unary operators have the highest precedence.
As the  <code>++</code> and <code>--</code> operators form
statements, not expressions, they fall
outside the operator hierarchy.

n = len(m.nat.limbs). func rr(m *Modulus) *Nat { rr := NewNat().ExpandFor(m) n := uint(len(rr.limbs)) mLen := uint(m.BitLen()) logR := _W * n // We start by computing R = 2^(_W * n) mod m. We can get pretty close, to // 2^⌊log₂m⌋, by setting the highest bit we can without having to reduce. rr.limbs[n-1] = 1 << ((mLen - 1) % _W) // Then we double until we reach 2^(_W * n). for i := mLen - 1; i < logR; i++ { rr.Add(rr, m) } // Next we need to get from R to 2^(_W * n) R mod m (aka from one to R in //...