I am a little bit confused about pointer index operator in C. I will try to explain my question with an example:

int array[5] = {1,2,3,4,5};
int *p;
p = array;
p[2]  ;

In the fourth line, I know that it increments the second index of array. However, when I see an index operator, I convert it.

For instance, I converted p[2] to *(p 2) . According to the operator precedence table, in the statement of *(p 2) , the increment and dereferencing operators have the same precedence, but increment takes precedence due to right associativity. Therefore, it becomes *(p 3). Then, this statement cannot change any value and just points third index of array.

Why does p[2] increment the second index of the array? What is wrong in my perspective?

CodePudding user response：

p[2] is equivalent to (*(p 2)) , not *(p 2) . You need an extra set of parentheses to maintain the precedence from the original expression.

Without them you've got *(p 2) which, as you've noted, is equivalent to *((p 2) ). This has a different meaning from the original expression since it splits up the 2 and the *. They need to be done in the same step since [2] is a single atomic operation.

CodePudding user response：

As already commented p[2] can be converted to (*(p 2)) , because p[2] is the element you want to increment. Think that when incrementing indexes is usually done like p[i ]