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

Because evaluation order is unspecified.

You are seeing nu in main being evaluated to 0 before even meth1 is called. This is the problem with chaining. I advise not doing it.

Just make a nice, simple, clear, easy-to-read, easy-to-understand program:

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