post-increment

Possible Duplicate: Behaviour and Order of evaluation in C# I have some code static void Main(string[] args) { int j = 0; for (int i = 0; i < 10; i++) j = j++; Console.WriteLine(j); } Why answer is 0 ? This is because of the way the ++ increment works . The order of operations is explained in this MSDN article This can be seen here (somebody please correct me if I am reading the spec on this one wrong :)): int j = 2; //Since these are value objects, these are two totally different objects now int intermediateValue = j; j = 2 + 1 //j is 3 at this point j = intermediateValue; //However j = 2 in

int main() { char arr[] = "geeksforgeeks"; char *ptr = arr; while(*ptr != '\0') ++*ptr++; printf("%s %s", arr, ptr); getchar(); return 0; } The statement inside while loop ++ptr++ behaves in a way that I don't understand. The post increment should have been evaluated first because of its high priority and the first output value should have been f(incrementing e). But that does not happen. To understand i changed the statement to ++*(ptr++) , so it may give the output what i expect(ffltgpshfflt is the output i expected;but actual output hffltgpshfflt). But still the output does not change. ()

Could you briefly explain me to understand this? Looking at the generated bytecode: public static void main(String[] args) { int x = 1; int y = 1; int z = 1; int a = 1; int b = 1; x = x + 1; y++; ++z; a += 1; b += 2; } generates (use javap -c classname ) 0: iconst_1 1: istore_1 2: iconst_1 3: istore_2 4: iconst_1 5: istore_3 6: iconst_1 7: istore 4 9: iconst_1 10: istore 5 12: iload_1 13: iconst_1 14: iadd 15: istore_1 16: iinc 2, 1 19: iinc 3, 1 22: iinc 4, 1 25: iinc 5, 2 28: return So using (jdk1.6.0_18): x = x + 1 creates 12: iload_1 13: iconst_1 14: iadd 15: istore_1 whereas y++; ++z; a +

This question already has an answer here: What is x after “x = x++”? 17 answers explain why difference in the same code [duplicate] 6 answers Below is my code snippet in C. void main(){ int x = 7; x = x++; printf("%d",x); } output : 8 public static void main(String[] args){ int x = 7; x = x++; System.out.println(x); } output : 7 i am not getting why both language giving different output. I've referred below link What is x after "x = x++"? Suresh Atta In java after x++ there is no change in the value of x x = x++; equal to int i= x; x = x + 1; x = i; so x remains same as i You can read more

This question already has an answer here: What is x after “x = x++”? 17 answers 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 ? NPE 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 has unspecified behaviour. See Undefined behavior and sequence points .) Hint: int b = 0, c; c = b++; c = b++; c = b++; c = b++;