c++ - 用户定义的文字可以有函数作为参数吗？

Question

函数可以与用户定义的文字一起使用吗？

如果是这样，可以做什么恶作剧？这合法吗？

void operator "" _bar(int (*func)(int)) {
  func(1);
}

int foo(int x) {
  std::cout << x << std::endl;
}

int main() {
  foo(0);    // print 0
  foo_bar;   // print 1
}

Answer 1

根据 C++11 Feb 2011 Draft § 2.14.8，用户文字类型是整数文字、浮点文字、字符串文字和字 rune 字。没有办法做函数文字类型。

A user-defined-literal is treated as a call to a literal operator or literal operator template (13.5.8). To determine the form of this call for a given user-defined-literal L with ud-suffix X, the literal-operator-id whose literal suffix identifier is X is looked up in the context of L using the rules for unqualified name lookup (3.4.1). Let S be the set of declarations found by this lookup. S shall not be empty.

整数:

operator "" X (n ULL)
operator "" X ("n")
operator "" X <’c1’, ’c2’, ... ’ck’>()

float :

operator "" X (f L)
operator "" X ("f")
operator "" X <’c1’, ’c2’, ... ’ck’>()

字符串:

operator "" X (str, len)
operator "" X <’c1’, ’c2’, ... ’ck’>() //unoffcial, a rumored GCC extension

字符:

operator "" X (ch)