我试图在一个表达式中重新实现Logan Smith的implicit_cast，我的方法是将一个匿名函数与type_identity_t结合起来：

#include <iostream>
using namespace std;
int main() {
  int i = 1;
  long j = 4;
  cout << max(j, []<class T>(type_identity_t<T> x) -> T { return x; }(i));
}

字符串
但它不起作用.编译器（gcc 13.2）抱怨说：

<source>: In function 'int main()':
<source>:6:70: error: no match for call to '(main()::<lambda(std::type_identity_t<T>)>) (int&)'
    6 |   cout << max(j, []<class T>(type_identity_t<T> x) -> T { return x; }(i));
      |                  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~
<source>:6:18: note: candidate: 'template<class T> main()::<lambda(std::type_identity_t<T>)>'
    6 |   cout << max(j, []<class T>(type_identity_t<T> x) -> T { return x; }(i));
      |                  ^
<source>:6:18: note:   template argument deduction/substitution failed:
<source>:6:70: note:   couldn't deduce template parameter 'T'
    6 |   cout << max(j, []<class T>(type_identity_t<T> x) -> T { return x; }(i));
      |                  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~

型
它是可以挽救的吗？还是我应该换个方法？