为什么赋值运算符不允许在声明对象的同一行中使用 lambda 表达式？
它似乎在 MSVC 中工作。
测试代码: https://godbolt.org/g/n2Tih1
class Func
{
typedef void(*func_type)();
func_type m_f;
public:
Func() {}
Func(func_type f) : m_f(f) {}
Func operator=(func_type f) {
m_f = f;
return *this;
}
};
int main()
{
// doesn't compile in GCC and clang, it does in MSVC
Func f1 = []() {
};
// compiles!
Func f2;
f2 = []() {
};
// compiles!
Func f3([]() {
});
}
最佳答案
Func f1 = []() {}; 为 copy initialization ，这需要两个用户定义的隐式转换来构造f1，第一个是从lambda到函数指针，第二个是从函数指针到Func。一个conversion sequence中只允许一个用户定义的隐式转换所以它失败了。
(强调我的)
If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T, or if T is non-class type, but the type of other is a class type, user-defined conversion sequences that can convert from the type of other to T (or to a type derived from T if T is a class type and a conversion function is available) are examined and the best one is selected through overload resolution.
和
Implicit conversion sequence consists of the following, in this order:
1) zero or one standard conversion sequence;
2) zero or one user-defined conversion;
3) zero or one standard conversion sequence.
对于 f2 = []() {}; 尝试调用适当的赋值运算符，Func 有一个并且它期望函数指针作为参数；只需一次从 lambda 到函数指针的隐式转换，就可以正常工作。
Func f3([]() {}); 为 direct initialization ，尝试调用适当的构造函数，Func 有一个，它期望函数指针作为参数。那么就和f2一样。
你可以从拷贝初始化和直接初始化的区别中理解。
In addition, the implicit conversion in copy-initialization must produce T directly from the initializer, while, e.g. direct-initialization expects an implicit conversion from the initializer to an argument of T's constructor.
关于c++ - 为什么在直接初始化和赋值中传递 lambda 而不是复制初始化时会编译？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50878474/