post-increment

I am trying to understand how pointer incrementing and dereferencing go together, and I did this to try it out: #include <stdio.h> int main(int argc, char *argv[]) { char *words[] = {"word1","word2"}; printf("%p\n",words); printf("%s\n",*words++); printf("%p\n",words); return 0; } I expected this code to do one of these: First dereference then increase the pointer (printing word1) First dereference then increase the value (printing ord1) Dereference pointer + 1 (printing word2) But compiler won't even compile this, and gives this error: lvalue required as increment operand am I doing something

I came across this problem in this website , and tried it in Eclipse but couldn't understand how exactly they are evaluated. int x = 3, y = 7, z = 4; x += x++ * x++ * x++; // gives x = 63 System.out.println(x); y = y * y++; System.out.println(y); // gives y = 49 z = z++ + z; System.out.println(z); // gives z = 9 According to a comment in the website, x += x++ * x++ * x++ resolves to x = x+((x+2)*(x+1)*x) which turns out to be true. I think I am missing something about this operator precedence. Java evaluates expressions left to right & according to their precedence. int x = 3, y = 7, z = 4; x

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

public void increment(){ int zero = 0; int oneA = zero++; // Compiles int oneB = 0++; // Doesn't compile int oneC = getInt()++; // Doesn't compile } private int getInt(){ return 0; } They are all int's, why won't B & C compile? Is it to do with the way ++ operator differs from = 0 + 1; ? Invalid argument to operation ++/-- i++ is an assignment to a variable i . In your case, zero++ is an equivalent to zero = zero + 1 . So 0++ would mean 0 = 0 + 1 , which makes no sense, as well as getInt() = getInt() + 1 . More accurately : int oneA = zero++; means int oneA = zero; zero = zero + 1; // OK, oneA

When does the post increment operator affect the increment? I have come across two opinions: 1) From http://gd.tuwien.ac.at/languages/c/programming-bbrown/c_015.htm : POST means do the operation after any assignment operation. 2) Closer home, an answer on SO (albeit on C++) says: ... that delays the increment until the end of the expression (next sequence point). So does the post increment operation... A) wait until a sequence point is reached or B) happen post an assignment operator or C) happen anytime before the sequence point? The correct interpretation is C, ie. the increment happens

Possible Duplicate: Undefined Behavior and Sequence Points In C++ on a machine code level, when does the postincrement++ operator get executed? The precedence table indicates that postfix++ operators are level 2: which means in int x = 0 ; int y = x++ + x++ ; // ans: y=0 The postfix ++'s execute first . However, it would seem that the logical operation of this line is the addition happens first (0+0), but how does that happen? What I imagine, is the following: // Option 1: // Perform x++ 2 times. // Each time you do x++, you change the value of x.. // but you "return" the old value of x there?