Increment operator on pointer of array errors?

问题

I'm trying something very simple, well supposed to be simple but it somehow is messing with me...

I am trying to understand the effect of ++ on arrays when treated as pointers and pointers when treated as arrays.

So,

int main()
{
    int a[4] = { 1, 4, 7, 9 };
    *a = 3;
    *(a+1) = 4;
    *++a = 4; //compiler error
}

1: So at *(a+1)=4 we set a[1]=4; //Happy But when *++a = 4;, I'd expect pointer a to be incremented one since ++ is precedent to * and then * kicks in and we make it equal to 4. But this code just does not work... Why is that?

Another problem:

int main()
{

    int* p = (int *)malloc(8);
    *p = 5;
    printf("%d", p[0]);

    *++p = 9; //now this works!
    printf("%d", p[1]); //garbage
    printf("%d", p[0]); //prints 9

}

2: Now *++p = 9; works fine but it's not really behaving like an array. How are two different? This is just incrementing p, and making it equal to 9. If I print p[0], it now prints 9 and I see that though can't access it via p[0] anymore, *(p-1) shows 5 is still there. So indexing a pointer with [0], where exactly does it point to? What has changed?

Thanks a lot all experts!

回答1:

The array names is not modifiable lvalue so operation ++ is not applied hence ++a that try to modify a is compilation time error (where a is array name).

Note *(a + 1) and *a++ are not same, a + 1 is a valid instruction as it just add 1 but doesn't modify a itself, Whereas ++a (that is equvilent to a = a + 1) try to modify a hence error.

Note 'array names' are not pointer. Pointers are variable but array names are not. Of-course when you assign array name to a pointer then in most expressions array names decays into address of first element. e.g.

int *p = a;

Note p points to first element of array (a[0]).

Read some exceptions where array name not decaying into a pointer to first element?

An expression a[i] is equivalent to *(a + i), where a can be either a pointer or an array name. Hence in your second example p[i] is valid expression.

Additionally, *++p is valid because because p is a pointer (a variable) in second code example.

回答2:

int a[4] = { 1, 4, 7, 9 };
int *pa=a;

There is one difference between an array name and a pointer that must be kept in mind. A pointer is a variable, sopa=a and pa++ are legal. But an array name is not a variable; constructions like a=pa and a++ are illegal