f()=X()??

[mahone]

Hey,

Please Correct me if I am wrong,when an expression contains '=' sign then the expression on the RHS of '=' sign is evaluated and the result is assigned to the variable or the identifier on the LHS of '=' sign, right?

But I am confused as what happens in case of two functions and an = sign.
Recently I found this Example in my book:

class X{
  int i;
  public:
   X(int ii=0);
   void modify();
};

X::X(int ii) {i=ii;}

void X::modify() {i++;}

X f5() {return X();}

const X f6() {return X();}

void f7(X& x) {x.modify();}

int main()
{
  f5()=X(1);
  f5().modify();
  //f7(f5());
  //f6()=X(1);
  //f6().modify();
  //f7(f6());
}

I have not ever seen some expression like this ever before:

f5()=X(1);

So can someone please tell me which of the above function will be called first f5() or X()? moreover is it a good programming practice to use two or more functions in a same expression and assigning the result to a result of third function?Also can someone please share with me some more examples in which two or more functions are used in a same expression like this.

Well, this is the first time that i have seen one function being equated to other and I am not liking it, this expression really feels so odd.Would be really helpful if someone can share more examples like this one so i can get along with thse kinds of statements.

Thank You for reading.

[Peter87]

The order of f5() and X(1) is not specified. It's up to the compiler to decide. The order doesn't make a difference in this case anyway.

f5() returns a temporary X object. X(1) also gives you a temporary object with i==1. Then operator= is used to copy the content so that both temporaries have i==1; Both objects are destroyed at the end of the line.

I know it doesn't look very useful but remember that operator= is a member function just like modify() is. You could have written line 20 as f5().operator=(X(1));. You can define operator= to do something else, like printing a message, but in that case it is probably better to name it something else.
f5().print(); <-- this make more sense?

[mahone]

so, is = a member function of every class by default?? Thanks for sharing this useful info.

[Peter87]

so, is = a member function of every class by default?

Yes.