将具有显式转义序列的字符串转换成相对字符 [英] Convert string with explicit escape sequence into relative character

问题描述

我需要一个函数来将显式转义序列转换为相对不可打印的字符。
Es：

I need a function to convert "explicit" escape sequences into the relative non-printable character. Es:

char str[] = "\\n";
cout << "Line1" << convert_esc(str) << "Line2" << endl:

会给出这个输出：

Line1

Line2

是有没有这样的功能？

推荐答案

我认为你必须自己编写这样的功能，因为转义字符是一个编译时间功能，即当您写入\\\
时，编译器将替换 \\\
序列与eol字符。生成的字符串为长度1 （不包括终止零字符）。

I think that you must write such function yourself since escape characters is a compile-time feature, i.e. when you write "\n" the compiler would replace the \n sequence with the eol character. The resulting string is of length 1 (excluding the terminating zero character).

在您的情况下，一个字符串\\ \\
是长度2 （再次不包括终止零），并包含 \ 和 n 。

In your case a string "\\n" is of length 2 (again excluding terminating zero) and contains \ and n.

您需要扫描字符串，遇到 \ 检查以下char。如果它是合法转义之一，您应该将其替换为相应的字符，否则跳过或将它们保持原样。

You need to scan your string and when encountering \ check the following char. if it is one of the legal escapes, you should replace both of them with the corresponding character, otherwise skip or leave them both as is.

（ http://ideone.com/BvcDE ）：

string unescape(const string& s)
{
  string res;
  string::const_iterator it = s.begin();
  while (it != s.end())
  {
    char c = *it++;
    if (c == '\\' && it != s.end())
    {
      switch (*it++) {
      case '\\': c = '\\'; break;
      case 'n': c = '\n'; break;
      case 't': c = '\t'; break;
      // all other escapes
      default: 
        // invalid escape sequence - skip it. alternatively you can copy it as is, throw an exception...
        continue;
      }
    }
    res += c;
  }

  return res;
}

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