pre-increment

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

问题 I did write a code to print all valid combinations of n-pair parentheses. However, at my first try, the algorithm output all the combinations twice, that is, . The code was: public static void solve(char[] string, int open, int closed, int index) { if (open < 0 || closed < open) { return; } if (open == 0 && closed == 0) { System.out.println(string); } if (open > 0) { string[index] = '('; solve(string, --open, closed, ++index); } if (closed > open) { string[index] = ')'; solve(string, open, -

问题 int i=1,j; j= ++i + ++i; printf("%d",j); The output of this program is 6 in C.But when I use the same logic for C#, the output is 5 . I want to know the reason why the same logic behaves differently in two different languages 回答1: The rule in C# is " evaluate each subexpression strictly left to right ". Therefore j= ++i + ++i ; is well defined in C# but the same expression invokes undefined behavior in C because you can't modify a variable more than once between two sequence points. C-FAQ:

问题 Salute.. I have an unusual problem. Here in this table in MSDN library we can see that precedence of () is higher than ++ (Pre-increment) . but when I run this code, it seems that precedence of ++(prefex) is higher: int main() { int a=3,b=2,x; x=++a + (a-b); cout<<"x= "<<x; return 0; } and the answer is : x=6 This happens with prefex ++ only and works as I expect with post-increment . Is there any reason? Regards.. int main() { int a=3,b=2,x; x=a++ + (a-b); cout<<"x= "<<x; return 0; } x=4 (I

问题 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 ? 回答1: 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

问题 I've just encountered a 'feature' in Javascript regarding pre-increments. In all other languages I've used, it goes like I thought it would. E.g. in C++: #include <iostream> int main() { int i = 0; i += ++i; std::cout << i << std::endl; // Outputs 2. } So, ++i doesn't make copy of the variable, hence the output is 2. Same in PHP: <?php $i = 0; $i += ++$i; echo $i; // Outputs 2. However, in Javascript: var i = 0; i += ++i; console.log(i); // Outputs 1. So it looks like that in Javascript, it

问题 This question already has answers here : Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed last year . I have seen in this site that prefix increment or postfix increment in a function call may cause undefined behavior. I have gone through one of those recently. The source code is something like this : #include <stdio.h> void call(int,int,int); int main() { int a=10; call(a,a++,++a); printf("****%d %d %d***_\n",a,a++,++a); return 0; } void call(int

问题 During a programming class, the professor was teaching us about x++ and ++x , with x being an integer. He said that in the scenario we are able to just put either x++ or ++x , ++x is more efficient (by little, but still, in theory, more efficient nonetheless). But I forgot why . Anyone knows? This was with Java. 回答1: It's not more efficient in Java. It can be more efficient in languages where the increment/decrement operators can be overloaded, but otherwise the performance is exactly the

问题 During a programming class, the professor was teaching us about x++ and ++x , with x being an integer. He said that in the scenario we are able to just put either x++ or ++x , ++x is more efficient (by little, but still, in theory, more efficient nonetheless). But I forgot why . Anyone knows? This was with Java. 回答1: It's not more efficient in Java. It can be more efficient in languages where the increment/decrement operators can be overloaded, but otherwise the performance is exactly the

问题 This question already has answers here : Closed 10 years ago . Exact Duplicate : Is there a performance difference between i++ and ++i in C++? Exact Duplicate : Difference between i++ and ++i in a loop? What is more efficient, i++ or ++i? I have only used this in Java and C/C++, but I am really asking for all languages that this is implemented in. In college I had a professor show us that ++i was more efficient, but it has been a couple of years, and I would like to get input from the Stack