scanf函数的正则表达式 - Ç [英] scanf regex - C

问题描述

我需要读取一个字符，直到按以下顺序写的是：\\ NX \\ N：

 （.....）\\ n
点¯x\\ n
 

\\ n是新行字符和（.....）可以是可以包括其它\\ n个字符的字符。

scanf函数使得普通的前pressions据我知道，但我不能使它读取字符串，直到这种模式。你能不能帮我scanf格式串中？

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我是想是这样的：

 字符输入[50000];
scanf函数（％[^（\\ NX \\ N）]，输入）;
 

，但它不工作

解决方案

  

scanf函数使得普通的前pressions据我所知

不幸的是，它不允许常规的前pressions：语法是误导性接近，但并没有什么甚至远程类似于 scanf函数的实施正则表达式。所有这一切的存在是的字符类的正则表达式的，所以支持％[＆LT;东西＆GT;] 被隐式视为 [＆LT;东西＆GT;] * 。这就是为什么你的 scanf函数的通话转化成阅读自由除其他字符'（'，'）'，'X'和'串\\ N'。

要解决手边的问题，你可以设置一个循环，读取字符输入字符。每当你收到了的'\\ n'，检查

• 您必须输入至少三个字符，你已经迄今所看到的，


• 该字符之前立即的'\\ n'是'X'和


• 这之前的字符在'X'是另一个的'\\ n'

如果所有上述是真实的，你已经达到了预期的输入序列的结束;否则，你的循环应该继续。

I needed to read a string until the following sequence is written: \nx\n :

(.....)\n
x\n

\n is the new line character and (.....) can be any characters that may include other \n characters.

scanf allows regular expressions as far as I know, but i can't make it to read a string untill this pattern. Can you help me with the scanf format string?

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I was trying something like:

char input[50000];
scanf(" %[^(\nx\n)]", input);

but it doesn't work.

解决方案

scanf allows regular expressions as far as I know

Unfortunately, it does not allow regular expressions: the syntax is misleadingly close, but there is nothing even remotely similar to the regex in the implementation of scanf. All that's there is a support for character classes of regex, so %[<something>] is treated implicitly as [<something>]*. That's why your call of scanf translates into read a string consisting of characters other than '(', ')', 'x', and '\n'.

To solve your problem at hand, you can set up a loop that read the input character by character. Every time you get a '\n', check that

• You have at least three characters in the input that you've seen so far,
• That the character immediately before '\n' is an 'x', and
• That the character before the 'x' is another '\n'

If all of the above is true, you have reached the end of your anticipated input sequence; otherwise, your loop should continue.

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