ruby – 我不明白为什么string.size返回它的作用

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

返回53.为什么？空白计数？即便如此.我们怎么得到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,因为％{case将每个/ n计为一个字符,并且在第一行之前认为是一个,最后一个,然后在第二行的末尾,而在EOS情况下只有一个在第1行之前和第1行之后？换句话说,为什么前者54和后者53？

解决方法

对于：

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<< EOS的情况下,其后面的行是字符串的一部分. <<<<<<与<<<<并且到该行的末尾将是“标记”的一部分,其确定字符串何时结束(在这种情况下,线上的EOS本身与<< EOS匹配). 在％{…}的情况下,它只是用来代替“…”的不同分隔符.因此,当你在％{之后的新行上开始字符串时,该换行符是字符串的一部分. 试试这个例子,您将看到％{…}与“…”的工作方式相同：

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