pre-increment

Half jokingly half serious : Why can't I do ++i++ in C-like languages, specifically in C#? I'd expect it to increment the value, use that in my expression, then increment again. Short answer: i++ is not an "lvalue", so can't be the subject of an assignment. Eric Lippert Though the short answer "it's not an lvalue" is correct , that's perhaps just begging the question. Why isn't it an lvalue ? Or, as we say in C#, a variable . The reason is because you cannot have your cake and eat it too . Work it out logically: First off, the meaning of a ++ operator in C#, whether postfix or prefix, is "take

I have two similar questions about operator precedences in Java. First one: int X = 10; System.out.println(X++ * ++X * X++); //it prints 1440 According to Oracle tutorial : postfix (expr++, expr--) operators have higher precedence than prefix (++expr, --expr) So, I suppose that evaluation order: 1) first postfix operator: X++ 1.a) X++ "replaced" by 10 1.b) X incremented by one: 10+1=11 At this step it should look like: System.out.println(10 * ++X * X++), X = 11; 2) second POSTfix operator: X++ 2.a) X++ "replaced" by 11 2.b) X incremented by one: 11+1=12 At this step it should look like: System

I understand the differences between i++ and ++i , but I'm not quite sure why I'm getting the results below: static void Main(string[] args) { int c = 42; c = c++; Console.WriteLine(c); //Output: 42 } In the above code, as this is assigning the variable to itself and then incrementing the value, I would expect the result to be 43 . However, it is returning 42 . I get the same result when using c = c--; as well. I realise I could just simply use c++; and be done with it, but I'm more curious why it is behaving the way that it is. Can anyone explain what's happening here? Let’s take a look at

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 variable ‘x’ is 5 then value of variable ‘b’ will be 5 because old value of ‘x’ is used. b = ++y; In this

Apparently on my Windows 8 laptop with HotSpot JDK 1.7.0_45 (with all compiler/VM options set to default), the below loop final int n = Integer.MAX_VALUE; int i = 0; while (++i < n) { } is at least 2 orders of magnitude faster (~10 ms vs. ~5000 ms) than: final int n = Integer.MAX_VALUE; int i = 0; while (i++ < n) { } I happened to notice this problem while writing a loop to evaluate another irrelevant performance issue. And the difference between ++i < n and i++ < n was huge enough to significantly influence the result. If we look at the bytecode, the loop body of the faster version is: iinc

Possible Duplicate: Is there a performance difference between i++ and ++i in C++? Is there a reason some programmers write ++i in a normal for loop instead of writing i++ ? For integers, there is no difference between pre- and post-increment. If i is an object of a non-trivial class, then ++i is generally preferred, because the object is modified and then evaluated, whereas i++ modifies after evaluation, so requires a copy to be made. ++i is slightly more efficient due to its semantics: ++i; // Fetch i, increment it, and return it i++; // Fetch i, copy it, increment i, return copy For int-like