如何在C ++中引用类结构中的函数？

Question

I'd like to use the getSth function which returns struct aa type in main() . Would you let me know the way to refer it?

//info.h
namespace nsp {
  class A {
    struct Info {
      struct aa {
        std::string str;
        int num;
        aa(): num(0) {}
      };
      std::vector<aa> aas;
      aa getSth();
    };
  };
}

//info.cpp
A::Info::aa A::Info::getSth() {
aa ret;
for(auto &tmp: aas) {
  if(ret.num < aas.num)
    ret.num = aas.num;
}
return ret;
}

// main.cpp
#include info.h
namepace nsp {
  class A;
}
int main()
{
  nsp::A *instance = new nsp::A();
  // How can I refer getSth using "instance"?
  .....
  return 0;
}

Answer 1

Quite simply put, you can't . You declared getSth inside a nested struct of type Info but didn't declare any data members of that type. So there's no object to call nsp::A::Info::getSth against.

What's worse, you declared A as a class and didn't provide an access specifier. A class's members are all private without an access specifier, so getSth can't be accessed outside the class. If you had done it this way instead:

class A {
  // other stuff; doesn't matter
  public:
  aa getSth();
};

well then, you could have accessed it from main like this:

int main()
{
  nsp::A *instance = new nsp::A();
  // now it's accessible
  instance->getSth();
  // deliberate memory leak to infuriate pedants
  return 0;
}