Encountered a strange statement in a book by C. So what is the address of an array variable?

Learn C in two textbooks Lately in the original and O'reilly's "Learning C programming" in the translation.
Faced with a strange statement, which may not properly understand (or "thank you" to Russian translation).

The words of the authors:
"The address of the array is the address of the array.
If you apply it to an array variable, the & operator, the result will not be different from the value of the variable.
&s == s."

Surprised. Checked:
#include <stdio.h>

int main ()
{
 char array[] = "Hello you!";
 char *pointer = "Hello you.";

 printf("This is an adress of array[]: %p\n\n", array);
 printf("This is an adress of array[]: %p\n\n", &array);
 printf("This is an adress of &array[1]: %p\n\n", &array[1]); // just To be sure
 printf("This is an adress of pointer: %p\n\n", pointer);

 return 0;
}


The result:
This is an adress of array[]: 0x7ffee5f78b6d

This is an adress of array[]: 0x7ffee5f78b6d

This is an adress of &array[1]: 0x7ffee5f78b6e

This is an adress of pointer: 0x109c87eb3


What & that no address of an array variable does not equal the value of the variable.

And yet, this is the vaunted O'reilly.
What's wrong here and whose is it?

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April 3rd 20 at 17:23
2 answers
April 3rd 20 at 17:25
array == &array[0]
&array is numerically equal to arraybut has a different type.
April 3rd 20 at 17:27
The array address is the address of the array. O'reilly write correctly.

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