This is what the standard says in 12.8 (Copying class objects). Copy construction: Each subobject is copied in the manner appropriate to its type: if the subobject is of class type, the copy constructor for the class is used; if the subobject is an array, each element is copied, in the manner appropriate to the … Read more
The standard correctly defines the return type of an assignment operator. Actually, the assignment operation itself doesn’t depend on the return value – that’s why the return type isn’t straightforward to understanding. The return type is important for chaining operations. Consider the following construction: a = b = c;. This should be equal to a … Read more
When you say let 5 = 10 it’s not a redefinition of 5, it’s a pattern matching, the same which occurs when you say foo 5 = undefined … foo 10 … The pattern simply fails if it’s ever matched. In let-expressions the match is lazy. This means the match is only being done when … Read more
[*] Your code causes undefined behavior. Not just “undefined if A is used as a base class and this, that or the other”. Actually undefined, always. return *this is already UB, because this is not guaranteed to refer to the new object. Specifically, consider 3.8/7: If, after the lifetime of an object has ended and … Read more
This is just guesswork on my part, but: 1. It’s hard to implement this in a single-pass compiler Lua’s bytecode compiler is implemented as a single-pass recursive descent parser that immediately generates code. It does not parse to a separate AST structure and then in a second pass convert that to bytecode. This forces some … Read more
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract: A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once. An example cited from §15.26.2 […] the following code is … Read more
Yes, they are both the exact same thing. They just take the reference of the object and reference it within the variable $o. Please note, thing should be variables.
A question that’s of greater importance in this case would be: “What does it mean, when a you write function in a way, that requires you to check for self assignment???” To answer my rhetorical question: It means that a well-designed assignment operator should not need to check for self-assignment. Assigning an object to itself … Read more