从 char* 创建包含转义字符的 char
考虑以下 char* 示例：
char* s = "\n";
如何将其转换为表示换行符的单个字符，如下所示：
char c = '\n';
除了处理换行符之外，我还需要能够将前面带有转义字符的任何字符转换为 char。这怎么可能？
Consider the following char* example:
char* s = "\n";
How can this be converted into a single char that represents the new line character like so:
char c = '\n';
In addition to processing newlines I also need to be able to convert any character with an escape character preceeding it into a char. How is this possible?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（4）
char c = *s;有效。字符串中的
'\n'在源码中只有两个字符：编译后是单个字符；对于所有其他转义字符也是如此。字符串
"fo\111\tar"编译后有 7 个字符（源代码中可见的 6 个字符（'f','o'< /code>、'\111'、'\t'、'a'和'r') 和一个空终止符)。char c = *s;works.The
'\n'inside the string is only two characters in source form: after compiling it is a single character; the same for all other escape character.The string
"fo\111\tar", after compiling, has 7 characters (the 6 visible in the source code ('f','o','\111','\t','a', and'r') and a null terminator).取消引用它：
Dereference it:
正如其他人所说，换行符实际上只是内存中的一个字符。要从字符串指针获取单个字符，您可以像访问数组一样访问指针（当然，假设为指向的字符串分配了内存）：
As others said, the newline is actually only one character in memory. To get a single character from a string pointer, you can access your pointer as if it is an array (assuming, of course, that there is memory allocated for the string pointed to):
你的意思是这样的吗？
Do you mean something like that?