postfix-operator

Defined this way, we can do neither ++x++ nor ++x-- . But on the other hand, both (++x)++ and (++x)-- are useful expressions: (++x)++ increments x by two and returns the value "in the middle", while (++x)-- is essentially equivalent to x+1 but completely avoids having to call operator+ , which can be quite useful sometimes. So why is the precedence not defined to have ++x++ automatically expand to (++x)++ rather than ++(x++) ? Is there some hidden meaning to the latter which I don't understand, or is it just to keep the precedence a simple list with all prefix operators making up one single

Currently I'm teaching a class of C++ programmers the basics of the C# language. As we discussed the topic operators I used C# standard categories of primary, unary etc. operators. One of the attendees felt puzzled, because in the C# standard the "postfix ++/--" have been put in the category of primary operators rather than the "prefix ++/--". Her rationale behind this confusion was, that she would rather implement the C++ operator "postfix ++/--" in terms of the operator "prefix ++/--". In other words she would rather count the operator "prefix ++/--" as a primary operator. - I understand her

By mistake, I wrote: ++number++; and got this: Uncaught ReferenceError: Invalid left-hand side expression in prefix operation Why? I would except this to to first increment number by one and then increment number by one again. In JavaScript, ++ is both the prefix and postfix increment operator. The postfix operator has higher precedence, and so when we apply precedence, your expression becomes: ++(number++); The result of number++ is a value, not a variable reference, and so it cannot be the operand of the prefix increment operator, for the same reason ++42 is invalid — there's nowhere to

I read from the official tutorial of Java that prefix and postfix ++ -- have different precedences: postfix: expr++ expr-- unary: ++expr --expr +expr -expr ~ ! Operators According to the tutorial, shouldn't this d = 1; System.out.println(d++ + ++d); print out 6 ( d++ makes d 2, ++d makes it 3) instead of 4? I know the explanation of ++d being evaluated beforehand, but if d++ has higher precedence then ++d , why isn't d++ being first evaluated? And what is more, in what case should d++ shows that it has higher precedence? EDIT: I tried the following: d = 1; System.out.println(++d * d++); It

please consider following code #include <iostream> using namespace std; class Digit { private: int m_digit; public: Digit(int ndigit=0){ m_digit=ndigit; } Digit& operator++();//prefix Digit& operator--(); //prefix Digit operator++(int); Digit operator--(int); int get() const { return m_digit;} }; Digit& Digit::operator++(){ ++m_digit; return *this; } Digit& Digit::operator--(){ --m_digit; return *this; } Digit Digit::operator++(int){ Digit cresult(m_digit); ++(*this); return cresult; } Digit Digit::operator--(int){ Digit cresult(m_digit); --(*this); return cresult; } int main(){ Digit cDigit(5

Possible Duplicate: What does a type followed by _t (underscore-t) represent? While typing in my IDE (Xcode), autocomplete pops up already-defined words when I'm partway thru entering some variable name. I occasionally see names that have ' _t ' at the end of them. What naming convention is that and what does it mean? Is there a reference document to look up pre- and post-fixes in common use? Searching with the term "postfix" gives me a lot of GoogleNoise about the mail server of the same name. The t stands for "type" or "typedef." You'll see a lot of POSIX headers (and others) with time_t ,

Have a look at these function signatures: class Number { public: Number& operator++ (); // prefix ++ Number operator++ (int); // postfix ++ }; Prefix doesn't take any parameter but postfix does. Why? I thought we can recognize them with different return types. Prefix and postfix ++ are different operators. With the standard Foo operator symbol(Foo &) style declaration there was no obvious way to distinguish the two. Rather than come up with some new syntax like Foo symbol operator(Foo &) which would make it into a special case unlike all the other operators and likely a bit of a pain to parse,