小编Voi*_*oid的帖子

class Song {

public:
    const string getAutherName();
}

void mtm::RadioManager::addSong(const Song& song,const Song& song1) {

    if (song.getAutherName() == song1.getAutherName())

}

我收到此错误:

Invalid arguments ' Candidates are: std::basic_string<char,std::char_traits<char>,std::allocator<char>> getAutherName() ' - passing 'const mtm::Song' as 'this' argument of 'std::string mtm::Song::getAutherName()' discards qualifiers [- fpermissive]

它为什么使用basic_string而不是string!怎么解决这个问题？

我在这里诽谤了一个结构

struct Owner{

char* ownerName;char* fatherName;char* address;};


void registerV(Owner *);

main(){

我在这里初始化所有者

struct Owner owner;

  owner.ownerName="Imran Ali";
  owner.fatherName="Ali Khokhar";
  owner.adress="KhushalPura";
  registerV(&owner);

}

在此功能中,我从用户那里获取输入,我必须使用其他功能在主要部分中显示.但是,当我尝试这样做时,我得到了垃圾值

void registerV(struct Owner *ownerPtr)
{

 char buyersName[50];
 char fatherName[50];
 char adress[100];

 cin.getline(buyersName, 50);
 cout << " Enter Buyers Name : " ;
 cin.getline(buyersName, 50);
 (*ownerPtr).ownerName=buyersName;

 cout << " Enter Fathers Name : " ;
 cin.getline(fatherName, 50);
 (*ownerPtr).fatherName=fatherName;

 cout << " Enter Adress : " ;
 cin.getline(adress, 100);
 (*ownerPtr).adress=adress;
}

当我尝试从主函数中查看值时,我得到垃圾值.请帮我.