const char **name VS char *name[]

Question

I know this topic was already discussed several times and I think I basically know the difference between arrays and pointer but I am interested in how arrays are exactly stored in mem.

for example:

const char **name = {{'a',0},{'b',0},{'c',0},0};
printf("Char: %c\n", name[0][0]); // This does not work

but if its declared like this:

const char *name[] = {"a","b","c"};
printf("Char: %c\n", name[0][0]); // Works well

everything works out fine.

Answer 1

When you define a variable like

char const*  str = "abc";
char const** name = &str;

it looks something like this:

+---+     +---+    +---+---+---+---+
| *-+---->| *-+--->| a | b | c | 0 |
+---+     +---+    +---+---+---+---+

When you define a variable using the form

char const* name[] = { "a", "b", "c" };

You have an array of pointers. This looks something like that:

          +---+     +---+---+
          | *-+---->| a | 0 |
          +---+     +---+---+
          | *-+---->| b | 0 |
          +---+     +---+---+
          | *-+---->| c | 0 |
          +---+     +---+---+

What may be confusing is that when you pass this array somewhere, it decays into a pointer and you got this:

+---+     +---+     +---+---+
| *-+---->| *-+---->| a | 0 |
+---+     +---+     +---+---+
          | *-+---->| b | 0 |
          +---+     +---+---+
          | *-+---->| c | 0 |
          +---+     +---+---+

That is, you get a pointer to the first element of the array. Incrementing this pointer moves on to the next element of the array.