在C ++ 11上模拟通用/模板化的lambda

Question

I got a problem on where I have something like this:

#define A_BODY printf("b is %s and c is %d, typeless! :3\n", b, c);
#define B_BODY return "test";
#define C_BODY return 42;

I need to make a code that would, for example, call a(b(), c()), with their respective types.

On C++14 I can easely do this:

template<typename B, typename C> auto a(B &&b, C &&c) {
  A_BODY
};
auto b() {
  B_BODY
};
auto c() {
  C_BODY
};

int main() {
  auto _b = b();
  auto _c = c();
  auto _a = a(_b, _c);

  return 0;
};

Achieving the desired result... is there any way to get the same result on C++11? :'(

Also, they can call each other recursively, so a simple ordering here wouldn't help either.

Edit

I will try to explain better my situation.

I got an input file like this, for example:

a is b c {
  printf("b is %s and c is %d\n", b, c);
  if(c > 42)
    printf("c is bigger than the truth!\n");
  return strlen(b) + c;
};
b is {
  return "test";
};
c is {
  return 42;
};

And I need to generate a C++ code that can properly call them. I'm trying to infer the return types without needing to define them on the input file.

From that, I can get the *_BODY specifications, the number of parameters, and the order of calls, but I don't know which type the parameters will be. So I'd need an templated lambda, for example, to do lazy evaluation of the function body.

I can do this with C++14 on GCC and CLang, but this is a commercial project and I need to support Visual Studio 2012 as well. I'm trying to find a workaround, if there is any. :(

Answer 1

Could do it like this:

#define A_EXPR printf("b is %s and c is %d, typeless! :3\n", b, c)
#define B_EXPR "test"
#define C_EXPR 42

template<typename B, typename C> auto a(B &&b, C &&c)
    -> decltype(A_EXPR) { return A_EXPR; }
auto b() -> decltype(B_EXPR) { return B_EXPR; }
auto c() -> decltype(C_EXPR) { return C_EXPR; }

int main() {
  auto _b = b();
  auto _c = c();
  auto _a = a(_b, _c);
  return 0;
};

Works fine in clang. On the other hand, gcc (4.8.1) complains that

sorry, unimplemented: string literal in function template signature

for function a , but you don't really need a result for that; it could be just void .