为什么指向int数组的未初始化指针可以分配变量的值？

Question

I have read answers to questions like why a pointer can be assigned value? however I'm still a bit confused about changing the values of pointers. Some of the comments don't seem to be completely accurate, or maybe it's implementation specific. (example: a[1] is exactly the same as writing *(a + 1). Obviously you can write *a = 3 since a is int*, so you can also write *(a + 1) = 3, so you can also write a[1] = 3 ).

Writing *a = 3 produces a warning: initialization makes pointer from integer without a cast. As well as a segfault.

My question is as follows.

int main(void)
{
    int b = 5, c = 10;
    int *a = b;

    *a = c; /* Will not work. (Should technically change the value of b to 10, leaving the pointer still pointing to b.) */
    printf("%d\n", *a);
    return 0;
}

The above example will not work, and will produce a segfault, but the following works for some reason I am not aware of.

int main(void)
{
    int a[10], i;
    for (i = 0; i < 10; ++i) {
        *(a + i) = i; /* Supposedly the same logic as '*a = c;', but works*/
    }

    for (i = 0; i < 10; ++i) {
        printf("%d\n", *(a + i));
    }

    return 0;
}

Thanks for your time and efforts.

**EDIT: Thank you for the answers, since it is *a = &b (I knew this (typo), but now the second example with the loop is unclear), the array indices are treated as variables, not as addresses I presume?

Answer 1

This:

int b = 5, c = 10;
int *a = b;

Doesn't work. You think the second line means this:

int *a;
*a = b;

But it doesn't. It means this:

int *a;
a = b;

The above error means that when you do this:

*a = c;

Your program crashes (or maybe not, undefined behavior!), because a is an invalid pointer (it points to address 5, which is wrong in many ways).