order of calling overloaded operators

[ES Kim]

iterator classes provide overloaded operators like this in general:

template <typename T>
class Iterator
{
Iterator operator++(int); // postfix ++
T& operator*();
};

Iterator<int> i;
*i++;

You know postfix ++ has higher precedence than unary *.
Does it gaurantee that postfix ++ is called before unary * operator?

 

[Jonathan Mcdougall]

iterator classes provide overloaded operators like this in general:

template <typename T>
class Iterator
{
Iterator operator++(int); // postfix ++
T& operator*();
};

Iterator<int> i;
*i++;

You know postfix ++ has higher precedence than unary *.

It`s the dereferencing operator, not unary * (which does not exist).

Does it gaurantee that postfix ++ is called before unary * operator?

Of course. The point of operator precedence is to guarantee the order
in which these operators are called.

*p++ is always *(p++) and a+b*c is always a+(b*c).


Jonathan

 

[Luke Meyers]

It's the dereferencing operator, not unary * (which does not exist).

Oh, come now. It's a unary operator represented by the symbol '*', and
you and everybody else knew exactly what the OP meant. In fact,
§5.3.1 [expr.unary.op] uses the exact term "the unary * operator" to
refer to this entity. Perhaps you just enjoy bullying, rather than
helping to inform? If you're going to be pedantic, have the decency to
be correct.

Luke

 

[Jonathan Mcdougall]

It's the dereferencing operator, not unary * (which does not exist).

Oh, come now. It's a unary operator represented by the symbol '*', and
you and everybody else knew exactly what the OP meant. In fact,
§5.3.1 [expr.unary.op] uses the exact term "the unary * operator" to
refer to this entity.

It was my intention to remove that line, but I forgot before posting. I
apologize.

Perhaps you just enjoy bullying, rather than
helping to inform? If you're going to be pedantic, have the decency to
be correct.

True.


Jonathan