Vin*_*ent 3 c++ compiler-warnings implicit-conversion narrowing c++14
请考虑以下代码,说明一些缩小的转换:
template <class T>
class wrapper
{
template <class> friend class wrapper;
public:
constexpr wrapper(T value)
: _data(value)
{}
template <class U>
constexpr wrapper(wrapper<U> other)
: _data(other._data)
{}
wrapper& operator=(T value)
{_data = value; return *this;}
template <class U>
wrapper& operator=(wrapper<U> other)
{_data = other._data; return *this;}
private:
T _data;
};
int main(int argc, char* argv[])
{
wrapper<unsigned char> wrapper1 = 5U;
wrapper<unsigned char> wrapper2{5U};
wrapper<unsigned char> wrapper3(5U);
wrapper<unsigned int> wrapper4 = 5U;
wrapper<unsigned int> wrapper5{5U};
wrapper<unsigned int> wrapper6(5U);
wrapper<unsigned char> wrapper7 = wrapper4; // Narrowing
wrapper<unsigned char> wrapper8{wrapper5}; // Narrowing
wrapper<unsigned char> wrapper9(wrapper6); // Narrowing
wrapper7 = wrapper4; // Narrowing
wrapper8 = wrapper5; // Narrowing
wrapper9 = wrapper6; // Narrowing
return 0;
}
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如何更改wrapper成员的主体,以便它触发编译器警告以缩小转换?我的目标是让用户意识到他们的代码可能存在问题.
您可以使用统一初始化语法触发缩小转换警告:
class wrapper
{
template <class> friend class wrapper;
public:
constexpr wrapper(T value)
: _data{value}
{}
template <class U>
constexpr wrapper(wrapper<U> other)
: _data{other._data} // note the curly brackets here
{}
wrapper& operator=(T value)
{_data = value; return *this;}
template <class U>
wrapper& operator=(wrapper<U> other)
{_data = {other._data}; return *this;} // and here
private:
T _data;
};
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同
wrapper<unsigned int> wrapper1 = 5U;
wrapper<unsigned char> wrapper2 = wrapper1; // Narrowing
wrapper<unsigned char> wrapper3(wrapper1); // Narrowing
wrapper<unsigned char> wrapper4{wrapper1}; // Narrowing
wrapper2 = wrapper1; // Narrowing
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最后四行中的任何一行都会在g ++中产生一个缩小的转换警告,以及来自clang中缩小转换的编译错误.