Basics of Memory Addresses in C
denniskubes.com
denniskubes.com
I remember in a programming language course we had to write a compiler for a subset of OCaml to MIPS assembly. I had literally no trouble adding support for references, while for others who did not grok C pointers (or had never programmed in C), it was less easy.
cl /FAsc source.c/source.cpp gives you assembly right there from the command line. For gcc: gcc -S
For dumping disassembly: DUMPBIN, or LINK /DUMP for MSVC, objdump -d/-D for binutils based (gcc, and others).
Debuggers: WinDBG (free), OllyDbg (Windows), gdb/ddd even WinGDB (not free)
First, learn assembly. Pick a machine architecture — it's not that important. For extra credit, pick multiple machine architectures and do something in each.
Then learn Lisp.
And Morse code.
You'll need Haskell for dat Damas-Milner sexiness.
Wait wait, we need to pile on some Ruby and JavaScript, you don't want to be some sort of weirdo that can't make a web app do you?
Hrrrrm. Yep, need some APL. With Unicode.
That'll really hammer in the concatenative into your brain.
And get some footing in OS design. Because, ultimately, your program's entire view of memory is just another API it uses to communicate with the OS.
It's like how everyone learns that (INT_MAX + 1) == INT_MIN (even non-developers seems to know this) yet that it actually is undefined (in C/C++), I feel that just noting that it isn't the whole truth (such as noting that the OS handles the memory behind your back) is quite valuable, even when learning the basics.
Otherwise you might end up feeling, as I do, that your foundation is shaky and built up on lies - not really knowing what "facts" you can trust.
Maybe you should explain the first element of an array having the same memory address as the actual array a little bit more, and relate it to why array indexing is 0-based too - the index is that many offsets from the beginning from the array.
The article says that arrays are different from pointers, but it does not prove it. It is quite simple to prove, see the program below.
Also, it's not interesting to limit the definition of arrays to just the locally and statically declared ones. If you do that then something like 90% of C programs (if not more, I think I never wrote such a C program except for exercises in class) don't use arrays at all. In all the other case (arrays passed as argument to a function, dynamically allocated arrays…), the are the same as pointers. Again, see the program below.
In reality, it is a bit pedantic to insist on this distinction, except for the rare case where it is a performance issue (the arrays of the article require one less memory access, the one to get the address of the memory at which the array starts).
#include <stdio.h>
#include <stdlib.h>
void
f (char a[], char *b, char *c)
{
printf("Once passed to a function as arguments:\n\n");
printf("What the article limits the definition of array to:\n");
printf("&a = %p\n", &a);
printf("a = %p\n", a);
printf("&(a[0]) = %p\n", &(a[0]));
printf("a + 1 = %p\n", a + 1);
printf("&(a[1]) = %p\n", &(a[1]));
printf("\n");
printf("Pointer to an array:\n");
printf("&b = %p\n", &b);
printf("b = %p\n", b);
printf("&(b[0]) = %p\n", &(b[0]));
printf("b + 1 = %p\n", b + 1);
printf("&(b[1]) = %p\n", &(b[1]));
printf("\n");
printf("Pointer to dynamically allocated memory:\n");
printf("&c = %p\n", &c);
printf("c = %p\n", c);
printf("&(c[0]) = %p\n", &(c[0]));
printf("c + 1 = %p\n", c + 1);
printf("&(c[1]) = %p\n", &(c[1]));
printf("\n");
}
int
main (int argc, char *argv[])
{
char a[4];
char *b = a;
char *c = malloc(sizeof(*c) * 4);
printf("What the article limits the definition of array to:\n");
printf("&a = %p\n", &a); /* behavior differs only here, this is the
difference with pointers */
printf("a = %p\n", a);
printf("&(a[0]) = %p\n", &(a[0]));
printf("a + 1 = %p\n", a + 1);
printf("&(a[1]) = %p\n", &(a[1]));
printf("\n");
printf("Pointer to an array:\n");
printf("&b = %p\n", &b);
printf("b = %p\n", b);
printf("&(b[0]) = %p\n", &(b[0]));
printf("b + 1 = %p\n", b + 1);
printf("&(b[1]) = %p\n", &(b[1]));
printf("\n");
printf("Pointer to dynamically allocated memory:\n");
printf("&c = %p\n", &c);
printf("c = %p\n", c);
printf("&(c[0]) = %p\n", &(c[0]));
printf("c + 1 = %p\n", c + 1);
printf("&(c[1]) = %p\n", &(c[1]));
printf("\n");
f(a, b, c);
return 0;
}
Here is a possible output of this program: What the article limits the definition of array to:
&a = 0x7fff5590ff00
a = 0x7fff5590ff00
&(a[0]) = 0x7fff5590ff00
a + 1 = 0x7fff5590ff01
&(a[1]) = 0x7fff5590ff01
Pointer to an array:
&b = 0x7fff5590fef8
b = 0x7fff5590ff00
&(b[0]) = 0x7fff5590ff00
b + 1 = 0x7fff5590ff01
&(b[1]) = 0x7fff5590ff01
Pointer to dynamically allocated memory:
&c = 0x7fff5590fef0
c = 0x202f010
&(c[0]) = 0x202f010
c + 1 = 0x202f011
&(c[1]) = 0x202f011
Once passed to a function as arguments:
What the article limits the definition of array to:
&a = 0x7fff5590fec8
a = 0x7fff5590ff00
&(a[0]) = 0x7fff5590ff00
a + 1 = 0x7fff5590ff01
&(a[1]) = 0x7fff5590ff01
Pointer to an array:
&b = 0x7fff5590fec0
b = 0x7fff5590ff00
&(b[0]) = 0x7fff5590ff00
b + 1 = 0x7fff5590ff01
&(b[1]) = 0x7fff5590ff01
Pointer to dynamically allocated memory:
&c = 0x7fff5590feb8
c = 0x202f010
&(c[0]) = 0x202f010
c + 1 = 0x202f011
&(c[1]) = 0x202f011
As we can see, for their practical use arrays and pointers can really be seen as the same thing. So again, except if you are optimizing a program where you can statically declare your arrays and access them a lot (i.e., you are doing matrix multiplication), the difference between arrays and pointers does not really matter.I posted a few comments on the site.