postfix-operator | 易学教程

Precedence of ++ and — operators in Java

阅读更多关于 Precedence of ++ and — operators in Java

问题 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

overloading postfix and prefix operators

阅读更多关于 overloading postfix and prefix operators

问题 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

How to differentiate (when overloading) between prefix and postfix forms of operator++? (C++)

阅读更多关于 How to differentiate (when overloading) between prefix and postfix forms of operator++? (C++)

Because I've overloaded the operator++ for an iterator class template<typename T> typename list<T>::iterator& list<T>::iterator::operator++() { //stuff } But when I try to do list<int>::iterator IT; IT++; I get a warning about there being no postifx ++ , using prefix form. How can I specifically overload the prefix/postifx forms? Daniel Earwicker Write a version of the same operator overload, but give it a parameter of type int . You don't have to do anything with that parameter's value. If you're interested in some history of how this syntax was arrived out, there's a snippet of it here .

i++ less efficient than ++i, how to show this?

阅读更多关于 i++ less efficient than ++i, how to show this?

I am trying to show by example that the prefix increment is more efficient than the postfix increment. In theory this makes sense: i++ needs to be able to return the unincremented original value and therefore store it, whereas ++i can return the incremented value without storing the previous value. But is there a good example to show this in practice? I tried the following code: int array[100]; int main() { for(int i = 0; i < sizeof(array)/sizeof(*array); i++) array[i] = 1; } I compiled it using gcc 4.4.0 like this: gcc -Wa,-adhls -O0 myfile.cpp I did this again, with the postfix increment

What does the postfix “_t” stand for in C? [duplicate]

阅读更多关于 What does the postfix “_t” stand for in C? [duplicate]

问题 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

What is the difference between prefix and postfix operators?

阅读更多关于 What is the difference between prefix and postfix operators?

问题 The following code prints a value of 9. Why? Here return(i++) will return a value of 11 and due to --i the value should be 10 itself, can anyone explain how this works? #include<stdio.h> main() { int i= fun(10); printf(\"%d\\n\",--i); } int fun (int i) { return(i++); } 回答1: There is a big difference between postfix and prefix versions of ++ . In the prefix version (i.e., ++i ), the value of i is incremented, and the value of the expression is the new value of i . In the postfix version (i.e.,

Why avoid increment (“++”) and decrement (“--”) operators in JavaScript?

阅读更多关于 Why avoid increment (“++”) and decrement (“--”) operators in JavaScript?

问题 One of the tips for jslint tool is: ++ and -- The ++ (increment) and -- (decrement) operators have been known to contribute to bad code by encouraging excessive trickiness. They are second only to faulty architecture in enabling to viruses and other security menaces. There is a plusplus option that prohibits the use of these operators. I know that PHP constructs like $foo[$bar++] has may easily result with off-by-one errors, but I couldn\'t figure out a better way to control the loop than a