class test {
public:
test &operator=(const test & other){} // 1
const test & operator+(const test& other) const {} // 2
const test & operator+(int m) {} //3
private:
int n;
};
int main()
{
test t1 , t2, t3;
// this works fine t1 = t2.operator+(t3) , calls 2 and 1
t1 = t2 + t3;
// this works fine t1 = t2.operator+ (100). calls 3 and 1
t1 = t2 + 100;
//this works fine t1 = (t2.operator+ (100)).operator+ (t3)
t1 = t2 + 100 +t3;
//why is the error in this one ????
// it should be t1 = (t2.operator+ (t3)).operator+ (100)
t1 = t2 + t3 + 100;
return 0;
}
答案 0 :(得分:8)
因为t2 + t3返回的对象是const,你不能调用它的非const函数(3)。
在“t1 = t2 + 100 + t3;”的情况下工作正常,因为t2 + 100返回的对象也是const,但是你正在调用它的const函数(2),这没关系。
答案 1 :(得分:1)
以下编译:
class test {
public:
test& operator=(const test & other){} // 1
test& operator+(const X& other) const {} // 2
test& operator+(int m) {} //3
private:
int n;
};
int main()
{
test t1, t2, t3;
t1 = t2 + t3;
t1 = t2 + 100;
t1 = t2 + 100 + t3; // <-- added a ';'
t1 = t2 + t3 + 100;
return 0;
}
答案 2 :(得分:1)
你错过了一个const:
const test & operator+(int m) /* ADD CONST HERE */ const {} //3