pre-increment

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

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

来源： https://stackoverflow.com/questions/25705000/the-difference-between-n-and-n-at-the-end-of-a-while-loop-ansi-c

来源： https://stackoverflow.com/questions/25705000/the-difference-between-n-and-n-at-the-end-of-a-while-loop-ansi-c

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; +

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

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

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

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

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&