memory-address

int t[10]; int * u = t; cout << t << " " << &t << endl; cout << u << " " << &u << endl; Output: 0045FB88 0045FB88 0045FB88 0045FB7C The output for u makes sense. I understand that t and &t[0] should have the same value, but how come &t is also the same? What does &t actually mean? When t is used on its own in the expression, an array-to-pointer conversion takes place, this produces a pointer to the first element of the array. When t is used as the argument of the & operator, no such conversion takes place. The & then explicitly takes the address of t (the array). &t is a pointer to the array

Please take a look at the picture below. When we create an object in java with the new keyword, we are getting a memory address from the OS. When we write out.println(objName) we can see a "special" string as output. My questions are: What is this output? If it is memory address which given by OS to us: a) How can I convert this string to binary? b) How can I get one integer variables address? Brian Agnew That is the class name and System.identityHashCode() separated by the '@' character. What the identity hash code represents is implementation-specific. It often is the initial memory address

In a reputable source about C, the following information is given after discussing the & operator: ... It's a bit unfortunate that the terminology [address of] remains, because it confuses those who don't know what addresses are about, and misleads those who do: thinking about pointers as if they were addresses usually leads to grief... Other materials I have read (from equally reputable sources, I would say) have always unabashedly referred to pointers and the & operator as giving memory addresses. I would love to keep searching for the actuality of the matter, but it is kind of difficult

In the book "Low-Level Programming: C, Assembly, and Program Execution on Intel® 64 Architecture" I read: Each virtual 64-bit address (e.g., ones we are using in our programs) consists of several fields. The address itself is in fact only 48 bits wide; it is sign-extended to a 64-bit canonical address. Its characteristic is that its 17 left bits are equal. If the condition is not satisfied, the address gets rejected immediately when used. Then 48 bits of virtual address are transformed into 52 bits of physical address with the help of special tables. Why is the difference in 4 bits between the