Question

I understand I did not make myself clear. My doubt, I think, could be summed up in this:

In an executable file(machine code) how are “variables” represented? Are they static memory addresses? Does the compiler gives each one a specific “name” (or just keeps the one you gave them)?

Expressed in code:

 int x=5;
 //Bunch of code
 cin>>y;
 cout<<x+1;

How does the program in each and every machine knows which address is going to hold the value 5, to hold the inputed value, to add 1 to the value it now holds and finally print that same value.

–João

Answer 1

Here is a simple program in C:

int main() {
    int a = 5;
    int b = 7;

    int c = a + b;

    return 0;
}

If you compile it with gcc -m32 -S -O0 -o main.s main.c under Linux, you’ll get something like this

    .file   "main.c"
    .text
    .globl  main
    .type   main, @function
main:
.LFB0:
    /* %ebp is a Base Pointer Register */
    pushl   %ebp
    movl    %esp, %ebp

    /* Here we reserve space for our variables */
    subl    $16, %esp

    /* a's address is %ebp - 4 */
    movl    $5, -4(%ebp)

    /* b's address is %ebp - 8 */
    movl    $7, -8(%ebp)

    /* a + b */
    movl    -8(%ebp), %eax
    movl    -4(%ebp), %edx
    addl    %edx, %eax

    /* c's address is %ebp - 12 */
    movl    %eax, -12(%ebp)

    /* return 0 */
    movl    $0, %eax
    leave
    ret

As you can see, in this case, variables’ addresses are calculated as offsets of a base pointer of a function. If you enable optimisations, variables’ values may be stored in registers.