C 和 C++ 中的指针

Question

I am exploring a concept of Pointers in C and C++. I have already read so many articles about it but still I have got some doubts. Looking forward that anyone of you can clear them.

So Here's my questions:

Declared Single Dimensional array like this:

int array[5] = {1, 2, 3, 4, 5};

Now I am printing below values:

printf("array--->%u", array);       // Output: 3199709072
printf("&array--->%u", &array);     // Output: 3199709072
printf("*array--->%d", *array);     // Output: 1
printf("*&array--->%d", *&array);   // Output: 3199709072

First three output are same as what I expected but I dont understand the fourth one.

Why is it printing an address of first element of array??

I am using *&array expression that means I am explicitly specifying that I want value at &array address but still it prints address of first element.

Any ideas??

Answer 1

When unary * and unary & comes together they nullify each others effect. *&array is equivalent to array . Since arrays decays to pointer to its first element when passed as an argument to a function, array will give address of the first element of the array array .

Looking it another way, the expression *&array will be parsed as (* (&array) ) ; retrieve the address of array array and then dereference it. Since &array is an address of an array, dereferencing it will give back the array itself, but as per the rule the dereferenced array will decay to pointer to its first element and it will make the expression *&array equivalent to array .

---

NOTE: Use %p specifier to print address. Like

printf("array--->%p", (void *)array);   

Otherwise, using wrong specifier will invoke undefined behavior.

Answer 2

Use the %p instead instead to get the pointer address in memory. Your code will generate undefined behaviors. Simply use the following example.

printf("%p \n",pointer);
//where pointer is array in your case

That should do.