A question

Question

I saw this question in a challenge:
int x=4
int *a=&x;
x--;
*a++;
cout<<x;//The answer is 3.
Why the changes made in pointer does not affect x?

Answer 1

Hi Joseph!

Good question!

This behaviour is explained by operator precedence.

*a++ is interpreted by the compiler as *(a++)

In effect the pointer is moved, and no longer has a reference to x.

If the code was ++*a; the compiler would see ++(*a) and x would be 4.

Answer 2

Thank you @jay