在C中引用char数组指针

Question

I ran into a pointer dereferencing problem.

In C, &a means the address of a. If a is a pointer ,then &a simply means the address of that pointer.

So if we have:

char ptr [] = "abcd"

ptr should be a pointer pointing to the first character, which is 'a'. therefore,

&ptr

should be the address of ptr, which is different than the address of 'a'. However, when I tried the following code I got really confused:

int main()
{
    char a [] = "abcd";

    printf("0x%X 0x%X", a, &a);

}

Output: 0xBF7E62AB 0xBF7E62AB

Can someone explain why a and &a have the same value? Based on my understanding they should be different. thanks in advance

Answer 1

So if we have: char ptr [] = "abcd" , ptr should be a pointer pointing to the first character.

No. Not at all.

ptr is an array. And an array is not a pointer.

Indeed, if you declared ptr as a real pointer, then you would get the expected behavior:

const char *ptr = "abcd";
printf("ptr = %p, &ptr = %p\n", (void *)ptr, (void *)&ptr);

As to why the address of the array is the same as the address of its first element: it's quite logical. The array represents a contiguous sequence of elements. The address of the array is where the array begins in memory. It begins where its first element begins. So, the address of the first element is (rather "can be" -- the standard does not mandate this behavior) the same as the address of the array itself.

+-----------+-----------+- - - -
| element 1 | element 2 |
+-----------+-----------+- - - -
^ start of array
^ start of first element

Answer 2

Can someone explain why a and &a have the same value? Based on my understanding they should be different.

In the statement

  printf("0x%X 0x%X", a, &a);  

Both a and &a are of different types. a is of char * type (after decay) while &a is of char (*)[5] type.
a decays to a pointer to the first element, therefore a is the address of first element of the string. While &a is the address of the string "abcd" and it is equal to the address of first element.

Answer 3

An array is not a pointer. That's right when you said &p is the address of the pointer, if p is definied like this:

char *p;

An array is different and don't has exactly the same behavior.

With the arrays:

char a[] = "abc"; , &a[0] is the address of the first element in your array, which is the same as a .

Answer 4

char a[] = "abcd" does not declare a pointer a to "abcd" but an array. Even if an array can decay to a pointer, it is a different type for which &a and a yield the same address.

Basically a yields the address to the first element of the array, so it is equivalent to &a (and to &a[0] ).

Answer 5

数组将始终引用其在内存中的存储位置，这就是为什么当您打印a时，它会为您提供它所在的地址，这等于使指针指向数组（＆a）

Answer 6

To get the behaveour you seek change

char a [] = "abcd";

to

char *a  = strdup("abcd");

or for a readonly string

const char *a  = "abcd";

You will then get a different address for a and &a.

When passing an array to a function, the array gets converted to a pointer. With your original program try

printf("%d %d\n",sizeof(a),sizof(&a));

The first will vary with the size of the string, the second will be based on the pointer size on your machine.