scanf 正则表达式 - C [英] scanf regex - C

问题描述

我需要读取一个字符串，直到写入以下序列: x :

(.....)
x

是换行符，(.....) 可以是任何可能包含其他 字符的字符.

就我所知，scanf 允许使用正则表达式，但在此模式之前我无法读取字符串.你能帮我扫描 scanf 格式字符串吗?

<小时>

我正在尝试类似的东西:

字符输入[50000];scanf(" %[^(
x
)]", 输入);

但它不起作用.

解决方案

scanf 据我所知允许使用正则表达式

不幸的是，它不允许使用正则表达式:语法接近于误导，但在 scanf 的实现中没有任何与正则表达式相似的东西.所有这些都支持正则表达式的字符类，所以%[<something>] 被隐式地视为 [.这就是为什么您对 scanf 的调用会转换为读取由 '(', ')'、'x' 和 ' ' 以外的字符组成的字符串.>

为了解决您手头的问题，您可以设置一个循环来逐个字符地读取输入.每次得到 ' ' 时，检查一下

• 您目前看到的输入中至少有三个字符，
• ' ' 之前的字符是 'x'，并且
• 'x'之前的字符是另一个' '

如果以上所有都为真，则您已到达预期输入序列的末尾；否则，您的循环应该继续.

I needed to read a string until the following sequence is written: x :

(.....)

x

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

---

I was trying something like:

char input[50000];
scanf(" %[^(
x
)]", 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 ' '.

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

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

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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