#include <initializer_list>
#include <iostream>
namespace {
class C {
public:
C(C const &) = delete;
C(C &&) = delete;
C(int) {
std::cout << "intn";
}
C(std::initializer_list<int>) {
std::cout << "initializern";
}
};
void f(C) {
}
// Compiles and prints "initializer" when called
C g() { return {0}; }
// Fails to compile
// C h() { return 0; }
} // namespace
int main() {
// Compiles and prints "initializer"
f({0});
// Fails to compile
// f(0);
}
能否在不调用initializer_list构造函数的情况下,将不可复制、不可移动的C类型构造为函数参数或函数返回值? 这只有在你可以改变C
以便选择所需的构造函数而不是初始化列表构造函数时才有可能,例如,通过将参数类型包装在不能转换为初始化列表构造函数的元素类型的东西中:
#include <initializer_list>
#include <iostream>
namespace {
template<class T>
struct wrap
{
T value;
};
class C {
public:
C(C const &) = delete;
C(C &&) = delete;
C(wrap<int>) {
std::cout << "intn";
}
C(std::initializer_list<int>) {
std::cout << "initializern";
}
};
void f(C) {
}
// Compiles and prints "int" when called
C g() { return {wrap<int>{0}}; }
} // namespace
int main() {
// Compiles and prints "int"
f({wrap<int>{0}});
g();
}
这个打印:
int
int