post-increment

I don't understand the results of following code: #include <stdio.h> #include <conio.h> int main() { int a[4]={1, 3, 5, 6}; //suppose a is stored at location 2010 printf("%d\n", a + 2); printf("%d", a++); return 0; } Why does the second printf function produce following error? error: lvalue required as increment operand Grijesh Chauhan Part-1: Array names are constant (not modifiable lvalue), your can add value to array name but can't modify it. Expression a + 2 doesn't modify a itself but when you do a++ that is equivalent to a = a + 1 try to modify array name --lvalue error. The expression a

My interest is in the difference between for and while loops. I know that the post-increment value is used and then incremented and the operation returns a constant pre-increment. while (true) { //... i++; int j = i; } Here, will j contain the old i or the post-incremented i at the end of the loop? zennehoy Since the statement i++ ends at the ; in your example, it makes no difference whether you use pre- or post-increment. The difference arises when you utilize the result: int j = i++; // i will contain i_old + 1, j will contain the i_old. Vs: int j = ++i; // i and j will both contain i_old +

I am a little confused about how the C# compiler handles pre- and post increments and decrements. When I code the following: int x = 4; x = x++ + ++x; x will have the value 10 afterwards. I think this is because the pre-increment sets x to 5 , which makes it 5+5 which evaluates to 10 . Then the post-increment will update x to 6 , but this value will not be used because then 10 will be assigned to x . But when I code: int x = 4; x = x-- - --x; then x will be 2 afterwards. Can anyone explain why this is the case? Sebastian Piu x-- will be 4, but will be 3 at the moment of --x , so it will end

I'm trying to get deeper with post and pre incrementors but am a bit stuck with the following expression : public static void main(String[] args) { int i = 0; i = i+=(++i + (i+=2 + --i) - ++i); // i = 0 + (++i + (i+=2 + --i) - ++i); // i = 0 + (1 + (3 + 2) - 1); // i = 0 + (6 - 1); System.out.println(i); // Prints 0 instead of 5 } I know I'm missing the logic somewhere but where? What I've tried : Going from left to right (though I know it is not recommended) Going from the insidest bracket and starting from there. Thanks for the help PS : The comments are the details of my calculus EDIT 1 I

This question already has an answer here: Why are these constructs using pre and post-increment undefined behavior? 14 answers I have some C code: main() { int a=1; void xyz(int,int); xyz(++a,a++); //which Unary Operator is executed first, ++a or a++? printf("%d",a); } void xyz(int x,int y) { printf("\n%d %d",x,y); } The function xyz has two parameters passed in, ++a and a++ . Can someone explain the sequence of operations to explain the result? The above code prints "3 13" or "2 23" depending on which compiler is used. Well, there are two things to consider with your example code: The order

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 Overflow community. i++ : create a temporary copy of i increment i return the temporary copy ++i : increment i return i With optimizations on, it is quite possible