// the lowest bucket was based around the 180ns/op figure for BenchmarkAccess,
		// plus some additional leeway to account for the apiserver doing other things
		// the largest bucket was chosen based on the fact that benchmarks indicate the
		// same Xeon running a CEL expression close to the estimated cost limit takes
		// around 760ms, so that bucket should only ever have the slowest CEL expressions

// For example, for deckSize=128 and handSize=4 this function might call pick(14); pick(73); pick(119); pick(26).
func (d *Dealer) Deal(hashValue uint64, pick func(int)) {
	// 15 is the largest possible value of handSize
	var remainders [15]int

	for i := 0; i < d.handSize; i++ {
		hashValueNext := hashValue / uint64(d.deckSize-i)
		remainders[i] = int(hashValue - uint64(d.deckSize-i)*hashValueNext)

#include <sys/cpuset.h>
#include <sys/param.h>
#include <sys/vdso.h>
*/
import "C"

// Local consts.
const (
	_NBBY            = C.NBBY            // Number of bits in a byte.
	_CTL_MAXNAME     = C.CTL_MAXNAME     // Largest number of components supported.
	_CPU_LEVEL_WHICH = C.CPU_LEVEL_WHICH // Actual mask/id for which.
	_CPU_WHICH_PID   = C.CPU_WHICH_PID   // Specifies a process id.
)

const (
	EINTR     = C.EINTR
	EFAULT    = C.EFAULT

#
# * types_${GOOS}.c
#
# This hand-written C file includes standard C headers and then
# creates typedef or enum names beginning with a dollar sign
# (use of $ in variable names is a gcc extension).  The hardest
# part about preparing this file is figuring out which headers to
# include and which symbols need to be #defined to get the
# actual data structures that pass through to the kernel system calls.

// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.

package math

func isOddInt(x float64) bool {
	if Abs(x) >= (1 << 53) {
		// 1 << 53 is the largest exact integer in the float64 format.
		// Any number outside this range will be truncated before the decimal point and therefore will always be
		// an even integer.

divides the drives you provide into erasure-coding sets of *2 to 16* drives.  Therefore, the number of drives you present must be a multiple of one of these numbers.  Each object is written to a single erasure-coding set.

Minio uses the largest possible EC set size which divides into the number of drives given. For example, *18 drives* are configured as *2 sets of 9 drives*, and *24 drives* are configured as *2 sets of 12 drives*.  This is true for scenarios when running MinIO as a standalone...

var tooLargeResourceVersionCauseMsg = "Too large resource version"

// NewTooLargeResourceVersionError returns a timeout error with the given retrySeconds for a request for
// a minimum resource version that is larger than the largest currently available resource version for a requested resource.
func NewTooLargeResourceVersionError(minimumResourceVersion, currentRevision uint64, retrySeconds int) error {

			//
			// This logic is equivalent to the logic in
			// cmd/compile/internal/types/size.go:calcStructOffset
			return 8
		}

		// spec: "For a variable x of struct type: unsafe.Alignof(x)
		// is the largest of the values unsafe.Alignof(x.f) for each
		// field f of x, but at least 1."
		max := int64(1)
		for _, f := range t.fields {
			if a := s.Alignof(f.typ); a > max {
				max = a
			}
		}
		return max

// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.

package zstd

import (
	"io"
	"math/bits"
)

// maxHuffmanBits is the largest possible Huffman table bits.
const maxHuffmanBits = 11

// readHuff reads Huffman table from data starting at off into table.
// Each entry in a Huffman table is a pair of bytes.

	x := new(big.Float).SetPrec(prec).SetInt64(1)

	// We use t as a temporary variable. There's no need to set its precision
	// since big.Float values with unset (== 0) precision automatically assume
	// the largest precision of the arguments when used as the result (receiver)
	// of a big.Float operation.
	t := new(big.Float)

	// Iterate.
	for i := 0; i <= steps; i++ {
		t.Quo(two, x)  // t = 2.0 / x_n