c++ - 用户定义的文字可以将函数作为参数吗？

Question

函数可以与用户定义的文字一起使用吗？

如果是这样，可以做些什么恶作剧？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、浮动文字、字符串文字和字符文字。没有办法做一个函数文字类型。

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.

Integers:

operator "" X (n ULL)
operator "" X ("n")
operator "" X <’c1’, ’c2’, ... ’ck’&gt;()

Floating:

operator "" X (f L)
operator "" X ("f")
operator "" X <’c1’, ’c2’, ... ’ck’&gt;()

String:

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

Character:

operator "" X (ch)

Answer 2

看foo_bar，它只是一个词法标记。它被解释为一个名为 的标识符foo_bar，而不是foo后缀_bar.

Answer 3

不。

C++ 有意避免这种恶作剧，因为如果符号 foo_bar 没有在您的示例中使用之前立即定义，则将很难理解。

您可以使用预处理器实现类似的效果。

#define bar (1)

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

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

Answer 4

I don't know if this adds anything but there's nothing preventing you from defining

PythonScript operator"" _python(const char*, std::size_t len) {...}

R"Py(
  print "Hello, World"
)Py"_python;

I actually think user-defined literals would make a nice way to embed scripts or SQL.