英文:
Visibility of function overloads with different scope
问题
下面的代码出错：error: ‘double& Sample::GetValue()’在此上下文中是受保护的 | double v = ms.GetValue();。如果删除受保护的函数，则可以编译通过。这是编译器的错误吗，不能选择具有不同范围的正确函数重载？
英文:
The code below gives error: error: ‘double& Sample::GetValue()’ is protected within this context | double v = ms.GetValue();. If the protected function if deleted it compiles fine. Is it the bug of the compiler that cannot select the correct function overload with different scope?
class Sample
{
public:
Sample() {}
virtual ~Sample() {}
const double& GetValue() const {return v;}
void Print() {std::cout << v << std::endl;}
protected:
double& GetValue() {return v;}
private:
double v;
};
class Sample2
{
public:
Sample2() {}
void Compute()
{
double v = ms.GetValue();
}
private:
Sample ms;
};
答案1
得分: 3
函数的可见性不会改变重载解析规则。这里没有错误，double& GetValue()只是比const double& GetValue() const更合适的候选项。
英文:
Visibility of a function doesn't alter overload resolution rules. There is no error here, double& GetValue() is simply a better candidate than const double& GetValue() const
答案2
得分: 1
梦之风说的是，但编译器不知道你想调用 const double& GetValue() const（顺便说一下，应该是 double GetValue() const），因为 ms 没有声明为 const。
不过有一个简单的解决办法：
void Compute()
{
const Sample &const_ms = ms;
double v = const_ms.GetValue();
}
英文:
What The Dreams Wind said, but the compiler doesn't know you want to call const double& GetValue() const (which should be double GetValue() const, btw) because ms is not declared const.
There's an easy fix though:
void Compute()
{
const Sample &const_ms = ms;
double v = const_ms.GetValue();
}
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