Question

我正在通过学习Visual C ++教科书来学习C ++ 当我想重载operator +时，我用来重载operator =的代码出错了。

#include <iostream>
#include <string.h>
using namespace std;
//This demo shows how default operator may cause conflict, so we use overloaded operator instead
class String{
    private:
        char* string;
        int len;
    public:
        String(const char*);
        String();
        ~String(){delete [] string;}
        String& operator=(String&);     //In the book it used //String & operator=(String&) but it went wrong
                                        //String object which returned by + only got value, not an initiated object
        String operator+(String&);      //Opreator +
        void show_string(){
            cout << "String: " << string << " \tString Address: " << (void*)string << " \tLength: " << len << endl;
        }
};

String::String():len(0){        //Constructor with no argument
    string = new char[len+1];
    string[0] = '\0';
}

String::String(const char* i_string):len(strlen(i_string)){ //Constructor
    string = new char[len+1];
    strcpy(string,i_string);
}

String& String::operator=(String& str_ref){     //Overloading operator =
//The function get reference and return itself
    delete [] string;
    cout << "Overloading Operator =...\n";
    len = str_ref.len;
    string = new char[len+1];
    strcpy(string,str_ref.string);
    return *this;
}

String String::operator+(String& str){
    cout << "Overloading Operator +...\n";
    char* strbuf = new char[len+str.len+1];
    strcpy(strbuf,string);
    strcat(strbuf,str.string);
    String retstr(strbuf);
    delete [] strbuf;
    return retstr;      //call by value coz we made a new String
}

int main(){
    String A_string("My ");
    String B_string("string"),C_string;
    cout << "Show (A_string+B_string)...\n";
    (A_string+B_string).show_string();
    C_string = A_string + B_string;
    cout << "Show C_string...\n";
    C_string.show_string();
return 0;
}

这很奇怪，因为它只在单独使用operator +或operator =时表现良好。

String A_string("Apple");
String B_string;
B_string = A_string;

(A_string+B_string).show_string();

这是错误

In function 'int main()':
56:11: error: no match for 'operator=' (operand types are 'String' and 'String')
  C_string = A_string + B_string;
           ^

note: candidate is:
note: String& String::operator=(String&)
 String& String::operator=(String& str_ref){  \\Overloading operator =
         ^
note:   no known conversion for argument 1 from 'String' to 'String&'

我以为我可以使用String作为参数String＆amp;，这在书中被告知所以我将operator =的参数更改为String，它可以工作。

String& operator=(String&);

到

String& operator=(String);

现在我很困惑何时只使用引用或字符串。

Answer 1

在本声明中

C_string = A_string + B_string;

由于执行运算符

，创建了String类型的临时对象

String operator+(String&);

您不能将非常量左值引用绑定到临时对象。

重写/添加赋值运算符，如

String& operator=( const String&);     
                   ^^^^

或添加移动赋值运算符。

String& operator=( String &&);     
                          ^^

C ++重载：[错误]与'operator ='不匹配（操作数类型为'String'和'String'）

1 个答案: