ruby - 我不明白为什么 string.size 返回它的作用

Question

long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS

返回 53。为什么？空格算不算？即便如此。我们如何得到 53？

这个怎么样？

     def test_flexible_quotes_can_handle_multiple_lines
    long_string = %{
It was the best of times,
It was the worst of times.
}
    assert_equal 54, long_string.size
  end

  def test_here_documents_can_also_handle_multiple_lines
    long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS
    assert_equal 53, long_string.size
  end

是这种情况吗，因为 %{ 案例将每个字符都计/n为一个字符，并且在第一行之前被认为是一个字符，在末尾一个，然后在第二行末尾，而在这种EOS情况下，在第一行之前只有一个行和第一行之后的一个？也就是说，为什么前者是54，而后者是53？

Answer 1

为了：

long_string = <<EOS
It was the best of times,
It was the worst of times.
EOS

String is:
"It was the best of times,\nIt was the worst of times.\n"

It was the best of times, => 25
<newline> => 1
It was the worst of times. => 26
<newline> => 1
Total = 25 + 1 + 26 + 1 = 53

和

long_string = %{
It was the best of times,
It was the worst of times.
}

String is:
"\nIt was the best of times,\nIt was the worst of times.\n"
#Note leading "\n"

这个怎么运作：

在 的情况下<<EOS，它后面的行是字符串的一部分。<<与行尾在同一行之后的所有文本<<都将成为确定字符串何时结束的“标记”的一部分（在这种情况下EOS，一行本身与 匹配<<EOS）。

在 的情况下%{...}，它只是用于代替 的不同分隔符"..."。因此，当字符串在 a 之后的新行开始时%{，该换行符是字符串的一部分。

试试这个例子，你会看到它%{...}是如何工作的"..."：

a = "
It was the best of times,
It was the worst of times.
"
a.length # => 54

b = "It was the best of times,
It was the worst of times.
"
b.length # => 53