#include <iostream>
#include <utility>
class test
{
private:
test() { }
public:
test foo() { return *this; }
static const char *name() { return "test"; }
};
int main()
{
std::cout << decltype(test().foo())::name() << std::endl; // 1
std::cout << decltype(std::declval<test>().foo())::name() << std::endl; // 2
}
我期望// 1行无法编译,因为test的默认构造函数是private.
However,it works well.我在我的g 4.8.3上测试了它–Wall -Wextra -Werror -pedantic难以置信,但它运行良好,没有任何错误或警告.
(此外,它似乎也适用于GCC 4.9.1.)
从this page开始,如果表达式未被评估,我想我们可以使用私有默认构造函数.所以,我测试了以下内容来检查它.
#include <iostream>
#include <utility>
class test
{
private:
test(int) { }
public:
test foo() { return *this; }
static const char *name() { return "test"; }
};
int main()
{
std::cout << decltype(test().foo())::name() << std::endl; // 1
std::cout << decltype(std::declval<test>().foo())::name() << std::endl; // 2
}
正如所料,它没有编译.
但为什么？？怎么可能？我们可以在未评估的表达中使用私人成员吗？或者默认构造函数是否有特殊规则？你能解释一下为什么吗？
解决方法
12.2/1 Even when the creation of the temporary object is unevaluated
or otherwise avoided,all the semantic restrictions shall be respected as if the temporary object had been created and later destroyed. [ Note: even if there is no call to the destructor or copy/move constructor,all the semantic restrictions,such as accessibility and whether the function is deleted,shall be satisfied. However,in the special case of a function call used as the operand of a decltype-specifier,no temporary is introduced,so the foregoing does not apply to the prvalue of any such function call. — end note ]
因此,即使未经评估,您仍然受限于创建和销毁临时所需的任何函数(包括构造函数)的可访问性.该注释的最后一句澄清了像declval这样的函数可以用来避免这个障碍.
