pre-increment

Pre and post increment in java [duplicate]

孤者浪人 提交于 2021-02-05 12:36:48
问题 This question already has answers here : How do the post increment (i++) and pre increment (++i) operators work in Java? (14 answers) Closed 2 years ago . I know how pre and post increment operators work, but recently I found out a strange behavior in Java. What i knew so far was (as explained in this answer as well): a = 5; i=++a + ++a + a++; => i=6 + 7 + 7; (a=8) which clearly shows that the ++a returns the value after increment, a++ returns the value before increment. But very recently, I

I'm having trouble understanding how Post Increment (++), Pre Increment work together in an example [duplicate]

浪子不回头ぞ 提交于 2020-12-23 18:28:33
问题 This question already has answers here : Undefined behavior and sequence points (5 answers) Behavior of post increment in cout [duplicate] (4 answers) Closed 13 days ago . I'm having trouble understanding how Post Increment (++), Pre Increment work together in an example. x++ means add 1 to the variable But I am confused with this example: using namespace std; / run this program using the console pauser or add your own getch, system("pause") or input loop */ int main() { int a; a=8; cout<<++a

pre-increment and post-increment in printf

╄→гoц情女王★ 提交于 2020-02-08 10:47:27
问题 int main() { int value = 4321; int *ptrVal = &value; printf("%d %d",++value,(*(int*)ptrVal)--); return 0; } How does pre-increment/post increment works in above print statement ? And why is answer 4321 4321 ? 回答1: You are modifying the object value twice between two sequence points: you are invoking undefined behavior. Undefined behavior means your program can print 4321 4321 , print 42 or even just crash. A correct version of your program would be: int value = 4321; int *ptrVal = &value; +

Can you have a incrementor and a decrementor on the same variable in the same statement in c

寵の児 提交于 2020-01-15 12:13:39
问题 Is --foo++; a valid statement in C? (Will it compile/run) And is there any practical application for this? Sorry for changing the question in an edit but I found something out. According to my C++ compiler (Visual Studio 2010): --++foo; is a valid command but foo--++; is not. Is there any reason for this? 回答1: No, it is not valid because the result of the increment / decrement operators is not a lvalue. EDIT: the OP edited his question by adding two more examples . So here we go, for the same

Can you have a incrementor and a decrementor on the same variable in the same statement in c

回眸只為那壹抹淺笑 提交于 2020-01-15 12:10:17
问题 Is --foo++; a valid statement in C? (Will it compile/run) And is there any practical application for this? Sorry for changing the question in an edit but I found something out. According to my C++ compiler (Visual Studio 2010): --++foo; is a valid command but foo--++; is not. Is there any reason for this? 回答1: No, it is not valid because the result of the increment / decrement operators is not a lvalue. EDIT: the OP edited his question by adding two more examples . So here we go, for the same

logical operation & pre-increment in c

不羁的心 提交于 2020-01-03 20:16:07
问题 Can any one explain why c still equal 15 after execution int main(void) { int t,a=5,b=10,c=15; t= ++a||++c; printf("%d %d %d",t,a,c); } 回答1: The logical-or operator || is a short-circuit operator. If the left side evaluates to a true boolean value (i.e. not 0), then the right side doesn't execute. Similarly for the logical-and operator && , if the left hand side is false (i.e. 0) the right hand side does not execute. 来源: https://stackoverflow.com/questions/31415375/logical-operation-pre

PHP Post and preincrement

微笑、不失礼 提交于 2020-01-03 17:26:37
问题 I have found in PHP some strange calculation, for example this: $c=5; $r = $c + ($c++ + ++$c); echo $r; Why result is 19 and not 17? Thanks 回答1: The result should be unspecified. Please read the following PHP specification: https://github.com/php/php-langspec/blob/master/spec/10-expressions.md While precedence, associativity, and grouping parentheses control the order in which operators are applied, they do not control the order of evaluation of the terms themselves. Unless stated explicitly

Mixed increment operators with logical operators [duplicate]

穿精又带淫゛_ 提交于 2019-12-29 01:25:49
问题 This question already has answers here : Is short-circuiting logical operators mandated? And evaluation order? (7 answers) Closed 2 years ago . I have a question concerning pre and post increments with logical operators if I have this code void main() {int i = - 3 , j = 2 , k = 0 , m ; m=++i||++j&&++k; printf("%d %d %d %d",i,j,k,m);} knowing that the increment and the decrement operators have higher precedence than && and || So they'll be executed first Then the && is higher than means -2||3&