postfix-operator | 易学教程

In Java, is a++ a postfix operator or an unary operator?

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问题 In Java, is a++ a postfix operator or an unary operator? In the Operator Precedence from Oracle, it is separated. Reference 回答1: It’s both. It’s a unary operator because it works solely on one term. If it’s x++ then it’s a postfix operator as it’s incremented after using the variable’s value. A prefix unary operator is ++x where it’s incremented before being used. Prefix operators have a higher precedence, but they’re both unary operators 来源： https://stackoverflow.com/questions/60014618/in

why can not we increment an array in similiar way as pointer in c? [duplicate]

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问题 This question already has answers here : Array increment operator in C (7 answers) Closed 4 years ago . #include <stdio.h> int main(){ int arr[] = {1, 2, 3, 4}; int *p; p = arr; printf("%d\n", *p); printf("%d\n", *arr); p++; printf("%d\n", *p); } This code outputs: 1 1 2 but when we add 2 lines as below: #include <stdio.h> int main(){ int arr[] = {1, 2, 3, 4}; int *p; p = arr; printf("%d\n", *p); printf("%d\n", *arr); p++; printf("%d\n", *p); arr++; printf("%d\n", *arr); } This code outputs:

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

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问题 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

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

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compilation order and post prefix opertors

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问题 I was wondering why the following outputs 7 7 6 7 instead of 5 6 6 7 my $a = 5; printf("%d %d %d %d",$a,++$a , $a++ , $a); I'm pretty sure it has something to do with the order of parameters compilation Thanks, 回答1: Before I start, let me point out that one should generally avoid situations where one you both sets and reads a variable within an expression. First, let's look at operand evaluation order. This isn't defined for many operators, but it is defined for the list operator. It's

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

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问题 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? 回答1: 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

c++ postfix / prefix operator overload as non-member function

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问题 I am writing my own array class as an exercise. Since, I read non-member functions are actually better in some ways than member functions. (Scott Meyers) I am trying to write as many operator overloads as non-member functions as possible. The operator overloads + , - all work out fine as non-member functions. my_array operator+(const my_array & left, const my_array & right); my_array operator-(const my_array & operand); my_array & operator++(); // prefix my_array operator++(int); //postfix,

Java: Prefix - Postfix issue

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问题 I have a small issue performing a subtraction on numbers using prefix and postfix operators. This is my program: public class postfixprefix { public static void main (String args[]) { int a = 5; int b; b = (a++) - (++a); System.out.println("B = " +b); } } Doing this, theoretically I am supposed to get 0 as the answer, but however, I am getting a -2. When I try to individually try to increment the values like in this program: public class postfixprefix { public static void main (String args[])

implementing a C++ postfix increment operator

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问题 I compiled the following example: #include <iostream> #include <iterator> using namespace std; class myiterator : public iterator<input_iterator_tag, int> { int* p; public: myiterator(int* x) :p(x) {} myiterator(const myiterator& mit) : p(mit.p) {} myiterator& operator++() {++p;return *this;} myiterator& operator++(int) {myiterator tmp(*this); operator++(); return tmp;} bool operator==(const myiterator& rhs) {return p==rhs.p;} bool operator!=(const myiterator& rhs) {return p!=rhs.p;} int&

compilation order and post prefix opertors

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I was wondering why the following outputs 7 7 6 7 instead of 5 6 6 7 my $a = 5; printf("%d %d %d %d",$a,++$a , $a++ , $a); I'm pretty sure it has something to do with the order of parameters compilation Thanks, Before I start, let me point out that one should generally avoid situations where one you both sets and reads a variable within an expression. First, let's look at operand evaluation order. This isn't defined for many operators, but it is defined for the list operator. It's documented to evaluate its operands in left-to-right order [1] . That means that printf 's arguments are evaluated