Difference between pointer to pointer and pointer

问题:

Given that the name of an array is actually a pointer to the first element of an array, the following code:

#include <stdio.h>

int main(void)
{
    int a[3] = {0, 1, 2};
    int *p;

    p = a;

    printf("%d\n", p[1]);

    return 0;
}

prints 1, as expected.

Now, given that I can create a pointer that points to a pointer, I wrote the following:

#include <stdio.h>                                                              

int main(void)                                                                  
{                                                                               
        int *p0;                                                                
        int **p1;                                                               
        int (*p2)[3];                                                           
        int a[3] = {0, 1, 2};                                                   

        p0 = a;                                                                 
        p1 = &a;                                                                
        p2 = &a;                                                                

        printf("p0[1] = %d\n(*p1)[1] = %d\n(*p2)[1] = %d\n",                    
                        p0[1], (*p1)[1], (*p2)[1]);                             

        return 0;                                                               
}

I expected it to compile and print

p0[1] = 1
(*p1)[1] = 1
(*p2)[1] = 1

But instead, it goes wrong at compile time, saying:

test.c: In function ‘main’:
test.c:11:5: warning: assignment from incompatible pointer type [enabled by default]

Why is that assignment wrong? If p1 is a pointer to a pointer to an int and a is a pointer to an int (because it's the name of an array of ints), why can't I assign &a to p1?

回答1:

Line 11 is

        p1 = &a;

where p1 has type int ** and a has type int[3], right?

Well; &a has type int(*)[3] and that type is not compatible with int** as the compiler told you

You may want to try

        p1 = &p0;

And read the c-faq, particularly section 6.

In short: arrays are not pointers, and pointers are not arrays.

回答2:

a is not a pointer to int, it decays to such in certain situations. If &a was of type int ** you couldn't very well use it to initialize p2, could you?

You need to do p1 = &p0; for the effect you want. "pointer to pointer" means "at this address, you will find a pointer". But if you look at the address &a, you find an array (obviously), so int ** is not the correct type.

回答3:

For many operations, a implies &a and both return the same thing: The address of the first item in the array.

You cannot get the address of the pointer because the variable does not store the pointer. a is not a pointer, even though it behaves like one in some cases.