删除char数组中的特殊字符和字符串

Question

我有一个类型为char array的电子邮件：

char email[80] = "pewdiepie@harvard.edu.au"

如何删除@后面的字符串并将pewdiepie作为最终结果？

Answer 1

例如以下方式

if ( char *p = std::strchr( email, '@' ) ) *p = '\0';

在C语言中，您必须在if语句之前声明变量p。例如

char *p;
if ( ( p = strchr( email, '@' ) ) != NULL ) *p = '\0';

如果使用类型为std::string的对象而不是字符数组，那么

std::string email( "pewdiepie@harvard.edu.au" );

然后你可以写

auto pos = email.find( '@' );
if ( pos != std::string::npos ) email.erase(pos);

Answer 2

简单地

#include <string>

std::string email("pewdiepie@harvard.edu.au");
email.erase(email.find('@'));

Answer 3

如果你坚持让它保持char []，而不是字符串

char email[80] = "pewdiepie@harvard.edu.au";
char name[80]; // to hold the new string

name[std::find(email, email + 80, '@') - email] = '\0'; // put '\0' in correct place
strncpy(name, email, std::find(email, email + 80, '@') - email); // copy the string

如果你想改变原来的char []，

*(std::find(email, email + 80, '@')) = '\0';