Does increment in pointer not same as that of integer?

Question

Consider the code below:

void increment(int* a)
{
    printf("%d\n",a);
    *a=*a+1;
}
int main()
{
    int a=10;
    int* p=&a;
    increment(&a);
    printf("%d",a);
    return 0;
}

This increments a to be 11 but this :

void increment(int* a)
{
    printf("%d\n",a);
     *a++;
}
int main()
{
    int a=10;
    int* p=&a;
    increment(&a);
    printf("%d",a);
    return 0;
}

The above code generates value as 10.

Is the pointer arithmetic not like integer arithmetic or am I missing something here??

Answer 1

The statement printf("%d\n",a); does not print the value of the integer pointed to by a. As written, it invokes undefined behavior.

printf("%p\n", (void*)a);

would print the value of the pointer, ie the address of the integer variable.

printf("%d\n", *a);

would print the value of the integer.

Futhermore, *a=*a+1; is not the same as *a++;. Due to operator precedence rules, *a++ is parsed as *(a++), the pointer is incremented, not the value pointed to. You could use ++*a; as a shorthand for *a = *a + 1;, and other variants are possible: ++a[0], a[0]++, (*a)++, but the preferred solution is:

*a += 1;