return false
}

// ScanWords is a split function for a [Scanner] that returns each
// space-separated word of text, with surrounding spaces deleted. It will
// never return an empty string. The definition of space is set by
// unicode.IsSpace.
func ScanWords(data []byte, atEOF bool) (advance int, token []byte, err error) {
	// Skip leading spaces.
	start := 0
	for width := 0; start < len(data); start += width {
		var r rune

		)

		// Total available space in the disk
		ch <- prometheus.MustNewConstMetric(
			prometheus.NewDesc(
				prometheus.BuildFQName(diskNamespace, "storage", "available"),
				"Total available space left on the drive",
				[]string{"disk"}, nil),
			prometheus.GaugeValue,
			float64(disk.AvailableSpace),
			disk.DrivePath,
		)

		// Total storage space of the disk
		ch <- prometheus.MustNewConstMetric(

		return l, true
	}
	return 0, false
}

// grow grows the buffer to guarantee space for n more bytes.
// It returns the index where bytes should be written.
// If the buffer can't grow it will panic with ErrTooLarge.
func (b *Buffer) grow(n int) int {
	m := b.Len()
	// If buffer is empty, reset to recover space.
	if m == 0 && b.off != 0 {
		b.Reset()
	}
	// Try to grow by means of a reslice.

	// A percentage value that indicates the size of the policy in packed form.
	// The service rejects any policy with a packed size greater than 100 percent,
	// which means the policy exceeded the allowed space.
	PackedPolicySize int `xml:",omitempty"`
}

// AssumeRoleWithWebIdentityResponse contains the result of successful AssumeRoleWithWebIdentity request.
type AssumeRoleWithWebIdentityResponse struct {

Тогда вы попробуете с давним другом программиста: автодополнением редактора.

Вы вводите первый параметр функции, `first_name`, затем точку (`.`), а затем нажимаете `Ctrl+Space`, чтобы запустить дополнение.

Но, к сожалению, ничего полезного не выходит:

<img src="/img/python-types/image01.png">

### Добавим типы

Давайте изменим одну строчку в предыдущей версии.

                + "                        f\n"
                + "g               g\n"
                + "h\\u0020h\n"
                + "\\   i=i\n"
                + "j=\\   j\n"
                + "space=\\   c\n"
                + "\n"
                + "dblbackslash=\\\\\n"
                + "                        \n";

        java.util.Properties props1 = new java.util.Properties();

import okio.ForwardingFileSystem
import okio.ForwardingSource
import okio.Path
import okio.Sink
import okio.Source
import okio.blackholeSink
import okio.buffer

/**
 * A cache that uses a bounded amount of space on a filesystem. Each cache entry has a string key
 * and a fixed number of values. Each key must match the regex `[a-z0-9_-]{1,64}`. Values are byte

   *
   * @since 8.0
   */
  public static final byte BEL = 7;

  /**
   * Backspace ('\b'): A format effector which controls the movement of the printing position one
   * printing space backward on the same printing line. (Applicable also to display devices.)
   *
   * @since 8.0
   */
  public static final byte BS = 8;

  /**

* Takes a `first_name` and `last_name`.
* Converts the first letter of each one to upper case with `title()`.
* <abbr title="Puts them together, as one. With the contents of one after the other.">Concatenates</abbr> them with a space in the middle.

{* ../../docs_src/python_types/tutorial001.py hl[2] *}

### Edit it

It's a very simple program.

But now imagine that you were writing it from scratch.

  }

  /**
   * Returns {@code n!}, that is, the product of the first {@code n} positive integers, or {@code 1}
   * if {@code n == 0}.
   *
   * <p><b>Warning:</b> the result takes <i>O(n log n)</i> space, so use cautiously.
   *
   * <p>This uses an efficient binary recursive algorithm to compute the factorial with balanced
   * multiplies. It also removes all the 2s from the intermediate products (shifting them back in at