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为什么我不能打电话给operator()?

[英]Why can't I call operator()?

I am trying to understand different topics in C++ by examples and I cannot get this example to work: 我试图通过示例理解C ++中的不同主题,但无法使该示例正常工作:

template<typename T>
class zero_init
{
    T val;
public:
    zero_init() : val(static_cast<T>(0)) { std::cout << "In constructor with no parameters\n"; }
    operator T() const { std::cout << "In operator T()\n";  return val; }
};

int main()
{
    const zero_init<int> x;
    x(); //Error!
    return 0;
}

I am obviously trying to call the operator() but it gives the error: "call of an object of a class type without appropriate operator()" 我显然正在尝试调用operator(),但它给出了错误:“在没有适当的operator()的情况下调用类类型的对象”

You accidentally implemented a type conversion operator and not operator() . 您不小心实现了类型转换运算符,而不是operator() Overload operator() like this instead (I removed the return value because you discard it in main anyway): 像这样重载operator() (我删除了返回值,因为无论如何您都将其丢弃在main ):

#include <iostream>

template<typename T>
class zero_init
{
    T val;
public:
    zero_init() : val(static_cast<T>(0)) { std::cout << "In constructor with no parameters\n"; }
    void operator()() const { std::cout << "In operator()\n"; }
};

int main()
{
    const zero_init<int> x;
    x();
    return 0;
}

If you actually need the return value, do it like this: 如果您确实需要返回值,请按照以下步骤操作:

#include <iostream>

template<typename T>
class zero_init
{
    T val;
public:
    zero_init() : val(static_cast<T>(0)) { std::cout << "In constructor with no parameters\n"; }
    T operator()() const { std::cout << "In operator()\n"; return val; }
};

int main()
{
    const zero_init<int> x;
    auto val = x();
    return 0;
}

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