Question

Sorry for this question but I am stuck. I have folowing syntax:

class xx
{
..some simple fields like: int t;  // )))
public: class  anotherClass;
xx();
MyObj* obj();
string*  name(); //error C2143: syntax error : missing  ;  before  *    

}

i have write # include <string> What does compiler wants from me?!

Answer 1

它希望你告诉他,string。你们想要的是标准一:

class xx
{
  public:
    std::string*  name();
};

Now, I m not sure why you would be returning a pointer to a string. That s a segmentation fault waiting to happen, if you ask me. Two m或e viable options that seem reasonable to me:

class xx
{
    std::string _name;
  public:
    const std::string& name() const 
    { 
        return _name; // WARNING: only valid as long as 
                      // this instance of xx is valid
    }
};

或

class xx
{
  public:
    std::string name() const { return "hello w或ld"; }
};

Answer 2

你们要么需要完全限定地加以扼杀,要么将其带入目前的名称空间:

std::string* name();

或

using std::string;

在一个头盔中,它通常认为污染全球名称空间的不良做法是可取的。

Answer 3

The compiler does not know what string is because string is residing in the namespace std, not in the global namespace. You need to change string to std::string.

在您的卷宗中,你可以使用“使用名空;”或“使用 st:扼杀”;然后是书写“扼杀”。但是,你永远不应在标题文件中使用名称空间声明。

BTW,因为其他人说,归还一块地str* 是不使用的,通常你会回座。

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