Two dimension char array initialization C

Question

According to this response pointers to pointers

char[] = char *p
char[][] = char **p

But, when I initialize an array of strings with the next two forms

char **p = {"Hello", "World"};
char p[][] = {"Hello", "World"};

The compiler shows me some Error.

[Error] array type has incomplete element type

But with

char *matriz[] = {"Hello", "World"};

There is no warning. I'm new in C language, and I'm very confused.

You cannot have dynamic-sized arrays in `C`, and that's not a warning, it is an ERROR — CinCout, Aug 29 '15 at 14:47
The answer in the link does not state what you are supposedly quoting!!! — barak manos, Aug 29 '15 at 14:51
What you quote ,states that `arrays and pointers are same` ? Is that what you meant? — ameyCU, Aug 29 '15 at 14:53
I understand that if I have `char arr[]`, then `arr = &arr[0]`, sorry but I'm very confused — rvillablanca, Aug 29 '15 at 14:57
wt is your compiler? Is it gcc? I'm not getting any error with gcc. http://goo.gl/SREx43 — ANjaNA, Aug 29 '15 at 15:10
With the pointer notation, I get same warnings as you, with `char p[][] = {"Hello", "World"};`, I get the error — rvillablanca, Aug 29 '15 at 15:14
@GargAnkit there is no dynamic array necessary; dimensions and sizes are statically known at compile time. — Peter - Reinstate Monica, Aug 29 '15 at 15:52
Any document that equates `char[][]` with `char **` is not going to be dreadfully helpful; the two are quite different. In fact, `char[]` is not the same as `char *` either, but when you use those notations in the parameter list for a function, they are equivalent (that is: `void something(char data[])` and `void something(char *data)` _are_ equivalent in this one important context). However, `void otherthing(char data[][10])` and `void otherthing(char **data)` are not equivalent at all. — Jonathan Leffler, Aug 29 '15 at 16:11
Having looked at the other SO question, I don't see where an answer makes the equation you infer from it — which is a relief since it is an SO question. — Jonathan Leffler, Aug 29 '15 at 16:18
The array of arrays `char ary[3][2]` decays to a pointer to the first element of the array, which will have the type `char (*)[2]` (pointer to array of char). No further decay can occur. This means `char **p;` and `char (*ary_ptr)[2];` are incompatible declarations, and `char **p;` and `char array[3][2];` are also incompatible declarations by extension of the concept. Whoever equated `char[][]` and `char**` is incorrect. — , Aug 29 '15 at 16:33

score 6 · Accepted Answer · answered Aug 29 '15 at 16:13

No an array of array is not the same as a double pointer. You really should read that up, there are a lot of FAQ and answers around.

Then the error that you receive is for the doubly empty [][]. In C, when you declare an array with [] the compiler tries to compute the size of the array from the initializer. This only works for one []. For the second you have to give a length. Something like

char p[][7] = {"Hello", "World"};

should work. For your first form

char **p

this declares a pointer to pointer, so in particular a pointer. Pointers can't be initialized with two items, they need just one, and that should itself be of the correct pointer type or convert to it.

Two dimension char array initialization C

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