Question

As far as I can understand, RAM is organized like a net of rows and columns of cells, each cell containing 1 byte. Also, each cell is label with an address memory written in hexadecimal system. Is this so? Now, when running a c++ program, I suppose it uses the RAM as a mean of storage. In this case, as the char type on c++ is the basic unit of storage, is this size of a char exactly the same as the cell (1 byte)?, does the size of a char depends on the size of a cell (in case the size of a cell is not 1 byte)?, does it depend on the compiler? Thank you so much.

Answer 1

It is easy to visualize RAM as a net of rows and columns. This is how most CS classes teach students as well and for most purposes this would do well at a conceptual level.
One thing you must know while writing C++ programs is the concept of 2 different memories: stack and heap. Stack is memory that stores variables when they come in scope. When they go out of scope, they are removed. Think of this as a stack implementation (FIFO).

Now, heap memory is slightly more complicated. This does not have anything to do with scope of the variable. You can set a fixed memory location to contain a particular value and it will stay there until you free it up. You can set the heap memory by using the ‘new’ keyword.
For instance: int* abc = new int(2);
This means that the pointer abc points to a heap location with the value ‘2’. You must explicitly free the memory using the delete keyword once you are done with this memory. Failure to do so would cause memory leaks.

In C, the type of a character constant like a is actually an int, with size of 4. In C++, the type is char, with size of 1. The size is NOT dependent on compiler. The size of int, float and the like are dependent on the configuration of your system (16/32/64-bit). Use the statement:

int a=5;
cout<<sizeof(a)<<endl;

to determine the size of int in your system.