This is a very simple question but even have some doubt in sequence point.
int a[3] = {1,2,4}; printf("%d",++a[1]); 3 Is this a valid c statement, I am getting output 3, which means it is same as
++(a[1]) But how is this possible as we have a pre-increment operator which has to increment the a first then the dereference has to happen.
Please correct my doubt. How we are getting 3?
Behavior is well defined. Operator [] has higher precedence than prefix ++ operator. Therefore operand a will bind to []. It will be interpreted as
printf("%d", ++(a[1])); Your parentheses are right, your rationale for what you think should happen obviously wrong.
If you were right, and prefix-increment had higher priority than indexing, you would get a compiler-error for ill-formed code, trying to increment an array.
As-is, there's absolutely no chance for sequencing-errors or the like leading to UB.
That's how pre increment operator works. Its similar to ++count. So here your value at a[1] (as [] has higher precendence than ++) get incremented and then its printed onto the console.
As you can see here: http://en.cppreference.com/w/cpp/language/operator_precedence
The operator [] has a higher precedence then ++
ais not even an assignable lvalue, so even if the precedence laws were different you could not get 4 . (you can't increment an array) (you cannot increment the numeric literal1either)