[英]C++ multiple conversion operators
I defined a class as below with two member variables and two conversion operators (operator float and operator chrono::microseconds).我用两个成员变量和两个转换运算符(运算符 float 和运算符 chrono::microseconds)定义了一个类,如下所示。 But the code below only works if I comment out the float operator.但是下面的代码只有在我注释掉浮点运算符时才有效。 Otherwise, it throws an error (cannot convert type).否则,它会抛出错误(无法转换类型)。 I can't figure out why?我想不通为什么?
#include <iostream>
#include <chrono>
using namespace std::chrono_literals;
class Peak {
public:
Peak (std::chrono::microseconds t, float magnitude)
: t_(t),
magnitude_(magnitude)
{
};
std::chrono::microseconds get_t() { return t_; }
//have to comment this out or I get an error
operator float() { return magnitude_; }
operator std::chrono::microseconds() {
return t_;
}
private:
std::chrono::microseconds t_{2us};
float magnitude_;
};
int main()
{
Peak a{3us, 100};
std::cout
<< "t is "
<< static_cast<std::chrono::microseconds>(a).count();
}
Looks like a compiler bug in GCC 7.3 and older.看起来像是 GCC 7.3 及更早版本中的编译器错误。 A workaround seems to be building using the -std=c++17
flag.解决方法似乎正在使用-std=c++17
标志构建。 Alternatively, making the float
conversion operator explicit
also fixes it:或者,使float
转换运算符explicit
也可以修复它:
explicit operator float() const { return magnitude_; }
(It's always a good idea to mark these const
, by the way.) (顺便说一下,标记这些const
总是一个好主意。)
Update:更新:
Actually, just making these operators const
seems to fix it already without the need to make the float
one explicit
nor to build with -std=c++17
.实际上,只需让这些运算符const
似乎已经修复它,而无需使float
explicit
也无需使用-std=c++17
构建。
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