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？空格很重要？甚至还是。我们如何获得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

发生这种情况是因为％{情况将每个/n都算作一个字符，并且在第一行之前被认为是一个字符，在第二行的末尾又是一个字符，而在EOS情况下则被认为是一个字符第一行，第一行之后？换句话说，为什么前者54和后者53？

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1 个回复

矾醒忻

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\"

怎么运行的：对于<<EOS，其后的行是字符串的一部分。 <<之后与<<在同一行上并到该行末尾的所有文本都是确定字符串何时结束的\“ marker \”的一部分（在这种情况下，一行上的EOS本身与<<EOS匹配）。在“ 11”的情况下，它只是代替“ 12”的分隔符。因此，当字符串在%{之后从新行开始时，该换行符就是字符串的一部分。尝试以下示例，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

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我不明白为什么string.size返回它的作用

1 个回复

发起人

ruby

heredoc

问题状态

我不明白为什么string.size返回它的作用

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1 个回复

发起人

ruby

heredoc

问题状态