pre-increment

i++ vs. ++i in a JavaScript for loop

由 我与影子孤独终老i 提交于 2019-12-09 17:10:22
问题 Because of JSLint, I almost always use i += 1 to increment a JavaScript for loop, but for quick and dirty scripts, I use i++ instead. However, I see a lot of for loops in other people's code in which they increment i by doing ++i instead. As far as I know, there is no difference in meaning between i++ and ++i , and jsPref shows no difference in performance. As such, I'm wondering where the convention of doing ++i comes from and why people tend to do it. Does anyone know why a lot of JS coders

Semantics of pre- and postfix “++” operator in Java [duplicate]

由 冷暖自知 提交于 2019-12-09 03:43:39
问题 This question already has answers here : Java: Prefix/postfix of increment/decrement operators? (10 answers) Closed 6 years ago . I wondering to know why this snippet of code give output 112 How this last digit 2 was creating? public static void main(String[] args) { int i = 0; System.out.print(++i); System.out.print(i++); System.out.print(i); Why does this happen? 回答1: Your snippet it's translated as int i = 0; i = i + 1; // 1 System.out.print(i); // 1 System.out.print(i); // 1 i = i + 1; //

Increment and logical operators precedence [duplicate]

由 两盒软妹~` 提交于 2019-12-09 01:50:32
问题 This question already has answers here : Short circuit behavior of logical expressions in C in this example (1 answer) Operator Precedence vs Order of Evaluation (6 answers) Closed 4 years ago . In the program shown below, prefix should be evaluated first because it has higher precedence, But answer is -2, 2, 0, 1 and it is explained in book "as LHS of || is true RHS is not evaluated." Why is it so? All the increments should performed first and then logical should be checked because of

post increment behaviour [duplicate]

由 大憨熊 提交于 2019-12-07 16:23:33
问题 This question already has answers here : What is x after “x = x++”? (17 answers) Closed 6 years ago . i have small doubt.why the below code is printing value i=2. int i=2; i=i++; System.out.println(i); can someone please explain me what is happening in line no 2. so there is no meaning here of doing ++ here? Thanks 回答1: i=i++; Because first the assignment happens, then the increment applies. Something like: first i gets 2, then ++ operation happens, but results won't be re-assigned to i, so i

pre-increment not working as i expect

由 点点圈 提交于 2019-12-07 15:54:05
问题 I am trying to learn dynamic programming by solving some questions online. One question that i came across requires the to process the following input 4 10 3 4 4 5 6 7 5 7 The first points at number of items, second the total capacity and the rest of the four (in a pair) should now point at the value and capacity. The problem i have is with the code which parses it. #include<iostream> #include<map> #include<iterator> using namespace std; template<typename T> void print(typename T::const

Behaviour of PreIncrement and PostIncrement operator in C and Java [duplicate]

由 ℡╲_俬逩灬. 提交于 2019-12-07 02:37:04
问题 This question already has answers here : Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed 6 years ago . I'm running the following programs in Visual C++ and Java: Visual C++ void main() { int i = 1, j; j = i++ + i++ + ++i; printf("%d\n",j); } Output: 6 Java: public class Increment { public static void main(String[] args) { int i = 1, j; j = i++ + i++ + ++i; System.out.println(j); } } Output: 7 Why the output in these two languages are different?

pre-increment not working as i expect

由 你说的曾经没有我的故事 提交于 2019-12-06 00:23:51
I am trying to learn dynamic programming by solving some questions online. One question that i came across requires the to process the following input 4 10 3 4 4 5 6 7 5 7 The first points at number of items, second the total capacity and the rest of the four (in a pair) should now point at the value and capacity. The problem i have is with the code which parses it. #include<iostream> #include<map> #include<iterator> using namespace std; template<typename T> void print(typename T::const_iterator start, typename T::const_iterator end) { typename T::const_iterator itr=start; while(itr!=end) {

The assignment to variable has no effect?

由 喜欢而已 提交于 2019-12-05 06:32:45
When I do this: count = ++count; Why do i get the warning - The assignment to variable count has no effect ? This means that count is incremented and then assigned to itself or something else ? Is it the same as just ++count ? What happens in count = count++; ? Why don't I get a warning for this ? count++ and ++count are both short for count=count+1 . The assignment is built in, so there's no point to assigning it again. The difference between count++ (also knows as postfix ) and ++count (also known as prefix ) is that ++count will happen before the rest of the line, and count++ will happen

What are the historical reasons C languages have pre-increments and post-increments?

由 时光毁灭记忆、已成空白 提交于 2019-12-03 17:15:16
问题 (Note: I am not asking about the definitions of pre-increment vs. post-increment, or how they are used in C/C++. Therefore, I do not think this is a duplicate question.) Developers of C (Dennis Ritchie et al) created increment and decrement operators for very good reasons. What I don't understand is why they decided to create the distinction of pre- vs post- increments/decrements? My sense is that these operators were far more useful when C was being developed than today. Most C/C++

What should be the output of echo ++$a + $a++ [duplicate]

由 孤人 提交于 2019-12-03 07:05:03
This question already has answers here : Why is $a + ++$a == 2? (13 answers) In the PHP manual, operator precedence section , there is this example: // mixing ++ and + produces undefined behavior $a = 1; echo ++$a + $a++; // may print 4 or 5 I understand the behavior is undefined because of the following reason: Since x + y = y + x the interpreter is free to evaluate x and y for addition in any order in order to optimize speed and/or memory. I concluded this after looking at the C code example in this article . My question is that the output of the above mentioned PHP code should be 4 no
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