C++ execution order in method chaining

Question

The output of this program:

#include  
class c1
{   
  public:
    c1& meth1(int* ar) {
      std::cout << \"method 1\" << std::e         


        

Answer 1

I think when compiling ,before the funtions meth1 and meth2 are really called, the paramaters have been passed to them. I mean when you use "c.meth1(&nu).meth2(nu);" the value nu = 0 have been passed to meth2, so it doesn't matter wether "nu" is changed latter.

you can try this:

#include  
class c1
{
public:
    c1& meth1(int* ar) {
        std::cout << "method 1" << std::endl;
        *ar = 1;
        return *this;
    }
    void meth2(int* ar)
    {
        std::cout << "method 2:" << *ar << std::endl;
    }
};

int main()
{
    c1 c;
    int nu = 0;
    c.meth1(&nu).meth2(&nu);
    getchar();
}

it will get the answer you want