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从正则表达式匹配中仅排除拖尾空格字符(Exclude only tailing space character from regex match)

 我试图从地址字符串中仅选择街道/单元号。  
 我的正则表达式是/^[\d\s,\-\/]*/  
 目前我的正则表达式在任何字母字符之前匹配digits '/' '-'和space characters字符。 但是我想排除拖尾space character 。  
 我的正则表达式匹配  
 悉尼新南威尔士州悉尼路61/2号（注意'2'后的空间特征）  
 代替  
 新南威尔士州曼利61/2悉尼路61/2号  
 
 以下是一些示例地址字符串：  
 61/2 Sydney Road, Manly NSW  
 61-2 Sydney Road, Manly NSW  
 61/2 3 Sydney Road, Manly NSW 

I am trying to select only the the street / unit number from an address string. 
My regex pattern is /^[\d\s,\-\/]*/ 
Currently my regex matches digits '/' '-' and space characters before any alphabetic characters. However I would like to exlude the tailing space character. 
My regex matches  
61/2Sydney Road, Manly NSW (note the space character after the '2') 
Instead of  
61/2 Sydney Road, Manly NSW 
 
Here are some sample addresses strings: 
61/2 Sydney Road, Manly NSW 
61-2 Sydney Road, Manly NSW 
61/2 3 Sydney Road, Manly NSW

原文：https://stackoverflow.com/questions/48738116

更新时间：2022-01-19 20:01

最满意答案

 当你这样做：  
char * str = "aaa bbb ccc ddd qqq";
 
 字符串文字"aaa bbb ccc ddd qqq"放在只读数据部分中，并使str指向它。 在这一点上，改变str指向的是未定义的行为，并希望提出一个段错误。  
 如果您绝对想要使用指针，请尝试以下方法：  
char * str = strdup("aaa bbb ccc ddd qqq");
 
 这将在堆上创建一个可写的字符串文字的副本 。  
 一旦你完成了它，你需要free(str); 释放记忆。 

When you do this: 
char * str = "aaa bbb ccc ddd qqq";
 
The string literal "aaa bbb ccc ddd qqq" is placed in the read-only data section, and str is made to point to it. At this point changing what str points to is undefined behavior and will hopefully raise a segfault. 
If you absolutely want to use a pointer, try this instead: 
char * str = strdup("aaa bbb ccc ddd qqq");
 
This will make a copy of the string literal on the heap, which will be writable. 
Once you're done with it though you'll need to free(str); to release the memory.

从正则表达式匹配中仅排除拖尾空格字符(Exclude only tailing space character from regex match)

最满意答案

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