post-increment

问题 Ruby does not support incrementing variables like variable++ . I saw this behaviour that: 2 ++ 4 gives 6 . In fact, any number of + signs between two variables are treated as one single + . How does ruby manage that? And since ruby does that, can it be taken as the reason for the unavailability of the ++ operator? 回答1: This: 2 ++ 4 is parsed as: 2 + (+4) so the second + is a unary plus. Adding more pluses just adds more unary + operators so: 2 ++++++ 4 is seen as: 2 + (+(+(+(+(+(+4)))))) If

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 This question already has answers here : Closed 7 years ago . Possible Duplicate: Pre & post increment operator behavior in C, C++, Java, & C# Here is a test case: void foo(int i, int j) { printf("%d %d", i, j); } ... test = 0; foo(test++, test); I would expect to get a "0 1" output, but I get "0 0" What gives?? 回答1: This is an example of unspecified behavior. The standard does not say what order arguments should be evaluated in. This is a compiler implementation decision. The compiler is

问题 Assume I have this code: int i = 2; int &ref = i++; Now, I understand that reference can not be initiaizied with rvalue, but I can not understand why ref isn't initialized with lvalue, meaning here, i , and after that i incremented. Moreover, in the following case: int i = 2; const int &ref = i++; cout << ref << endl; 2 will be printed, and it means that ref initialized with i before increment, i.e. initialized with lval. After that ref is incremented, but ref is const. Can someone explain me

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 This question already has answers here : What is x after “x = x++”? (17 answers) Closed 5 years ago . I was asked in an interview the below question. int b = 0; b = b++; b = b++; b = b++; b = b++; what will be the value of b after every line execution ? The output is coming as 0 for every line. Why is the output not coming as 0,1,2,3 ? 回答1: In Java, the expression b = b++ is equivalent to int tmp = b; b = b + 1; b = tmp; Hence the result. (In some other languages, the exact same expression

问题 Why does this int x = 2; for (int y =2; y>0;y--){ System.out.println(x + " "+ y + " "); x++; } prints the same as this? int x = 2; for (int y =2; y>0;--y){ System.out.println(x + " "+ y + " "); x++; } As far, as I understand a post-increment is first used "as it is" then incremented. Are pre-increment is first added and then used. Why this doesn't apply to the body of a for loop? 回答1: The loop is equivalent to: int x = 2; { int y = 2; while (y > 0) { System.out.println(x + " "+ y + " "); x++;

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 I am learning programming and I have started from C language. I was reading Let us C book. And I was going through this program in that book. main( ) { int a[5] = { 5, 1, 15, 20, 25 } ; int i, j, k = 1, m ; i = ++a[1] ; j = a[1]++ ; m = a[i++] ; printf ( "\n%d %d %d", i, j, m ) ; } My understanding was, it will print i as 2 , j as 1 and m as 15 But somehow it is printing as i as 3 , j as 2 and m as 15 ? Why is it so? Below is my understanding- b = x++; In this example suppose the value of