Let us assume,
int *p;
int a = 100;
p = &a;
What will the following code actually do and how?
p++;
++p;
++*p;
++(*p);
++*(p);
*p++;
(*p)++;
*(p)++;
*++p;
*(++p);
I know, this is kind of messy in terms of coding, but I want to know what will actually happen when we code like this.
Note : Lets assume that the address of a=5120300, it is stored in pointer p whose address is 3560200. Now, what will be the value of p & a after the execution of each statement?
First, the ++ operator takes precedence over the * operator, and the () operators take precedence over everything else.
Second, the ++number operator is the same as the number++ operator if you're not assigning them to anything. The difference is number++ returns number and then increments number, and ++number increments first and then returns it.
Third, by increasing the value of a pointer, you're incrementing it by the sizeof its contents, that is you're incrementing it as if you were iterating in an array.
So, to sum it all up:
ptr++; // Pointer moves to the next int position (as if it was an array)
++ptr; // Pointer moves to the next int position (as if it was an array)
++*ptr; // The value pointed at by ptr is incremented
++(*ptr); // The value pointed at by ptr is incremented
++*(ptr); // The value pointed at by ptr is incremented
*ptr++; // Pointer moves to the next int position (as if it was an array). But returns the old content
(*ptr)++; // The value pointed at by ptr is incremented
*(ptr)++; // Pointer moves to the next int position (as if it was an array). But returns the old content
*++ptr; // Pointer moves to the next int position, and then gets accessed, with your code, segfault
*(++ptr); // Pointer moves to the next int position, and then gets accessed, with your code, segfault
As there are a lot of cases in here, I might have made some mistake, please correct me if I'm wrong.
EDIT:
So I was wrong, the precedence is a little more complicated than what I wrote, view it here: http://en.cppreference.com/w/cpp/language/operator_precedence