C++操作员错误
收到此错误：
C:\CodeBlocks\kool\praks3\vector.h|62|错误：传递 'const Vector<2u>'作为 'std::string Vector::toString() [with Short unsigned int n = 2u]' 的 'this' 参数丢弃限定符|
用这个代码：
#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <sstream>
template <unsigned short n>
class Vector {
public:
std::vector<float> coords;
Vector();
Vector(std::vector<float> crds);
float distanceFrom(Vector<n> v);
std::string toString();
template <unsigned short m>
friend std::ostream& operator <<(std::ostream& out, const Vector<m>& v);
};
template <unsigned short n>
std::string Vector<n>::toString() {
std::ostringstream oss;
oss << "(";
for(int i = 0; i < n; i++) {
oss << coords[i];
if(i != n - 1) oss << ", ";
}
oss << ")";
return oss.str();
}
template <unsigned short m>
std::ostream& operator<<(std::ostream& out, const Vector<m>& v) {
out << v.toString(); // ERROR HEEEERE
return out;
}
getting this error:
C:\CodeBlocks\kool\praks3\vector.h|62|error: passing 'const Vector<2u>' as 'this' argument of 'std::string Vector::toString() [with short unsigned int n = 2u]' discards qualifiers|
with this code:
#include <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <sstream>
template <unsigned short n>
class Vector {
public:
std::vector<float> coords;
Vector();
Vector(std::vector<float> crds);
float distanceFrom(Vector<n> v);
std::string toString();
template <unsigned short m>
friend std::ostream& operator <<(std::ostream& out, const Vector<m>& v);
};
template <unsigned short n>
std::string Vector<n>::toString() {
std::ostringstream oss;
oss << "(";
for(int i = 0; i < n; i++) {
oss << coords[i];
if(i != n - 1) oss << ", ";
}
oss << ")";
return oss.str();
}
template <unsigned short m>
std::ostream& operator<<(std::ostream& out, const Vector<m>& v) {
out << v.toString(); // ERROR HEEEERE
return out;
}
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（3）
将
toString方法设为const：这
是因为您将
const限定符添加到中的。您只能对Vector运算符<<const限定对象调用const限定方法。Make the
toStringmethodconst:and
That's because you add the
constqualifier to theVectorin theoperator<<. You are only allowed to callconstqualified methods onconstqualified objects.这是因为您的向量被声明为 const，而您的 toString 运算符不是 const 方法。
因此，禁止使用 const 实例调用此方法。
如果在将向量转换为字符串时不对其进行编辑，则应将其声明为 const 方法：
This is because your vector is declared as const, while your toString operator is not a const method.
Therefore, calling this method is forbidden with a const instant.
If you do not edit the vector while converting it to string, you should declare it as a const method :
如果您有一个
const Foo，则只能调用它的const成员函数。因此将其声明为std::string toString() const。If you have a
const Foo, you can only invokeconstmember functions on it. So declare it asstd::string toString() const.