C语言 在 C 字符串中包含双引号 (")
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Include double-quote (") in C-string
提问by Dragos Rizescu
I would like a to define a variable string in C that contains the following set of characters: a-zA-Z0-9'-_”.
我想在 C 中定义一个包含以下字符集的变量字符串：a-zA-Z0-9'-_”.
Therefore I would do it like this:
因此，我会这样做：
char str[64] = "abcdefghijklmnopqrstuwxyzABCDEFGHIJKLMNOPQRSTUWXYZ0123456789'-_""
As you can see the problem is at the end with "character.
如您所见，问题在"字符的末尾。
Question 1:How can I work around that?
问题 1：我该如何解决这个问题？
Question 2:Is there a better way than my way to define a string like that?
问题 2：有没有比我定义这样的字符串更好的方法？
PS:I didn't really know how to title my question, so if you got a better one, please edit it.
PS：我真的不知道如何命名我的问题，所以如果你有更好的，请编辑它。
回答by Glenn Teitelbaum
use the backslash: "\""is a string containing "
使用反斜杠："\""是一个包含"
like this:
像这样：
char str[67] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'-_\"";
added one for the implicit '\0' at the end (and put in the missing vV) - this could also be:
最后为隐式 '\0' 添加了一个（并放入丢失的 vV） - 这也可能是：
char str[] = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'-_\"";
and let the compiler count for you - then you can get the count with sizeof(str);
并让编译器为您计数 - 然后您可以使用sizeof(str);
How does it add up to 67?
它如何加起来为 67？
a-z 26
A-Z 26
0-9 10
'-_" 4
'##代码##' 1
---
67
回答by Bilal Syed Hussain
Use "\""(backslash") for putting "in a string
使用"\""(backslash") 放入"字符串
