C compound literals, pointer to arrays

I'm try开发者_如何学运维ing to assign a compound literal to a variable, but it seems not to work, see:

  int *p[] = (int *[]) {{1,2,3},{4,5,6}};

I got a error in gcc.

but if I write only this:

  int p[] = (int []) {1,2,3,4,5,6};

Then it's okay.

But is not what I want.

I don't understand why the error occurrs, because if I initialize it like a array, or use it with a pointer of arrays of chars, its okay, see:

  int *p[] = (int *[]) {{1,2,3},{4,5,6}}; //I got a error
  int p[][3] = {{1,2,3},{4,5,6}}; //it's okay
  char *p[] = (char *[]) {"one", "two"...}; // it's okay!

Note I don't understand why I got an error in the first one, and please I can't, or I don't want to write like the second form because it's needs to be a compound literals, and I don't want to say how big is the array to the compiler. I want something like the second one, but for int values.

Thanks in advance.

First, the casts are redundant in all of your examples and can be removed. Secondly, you are using the syntax for initializing a multidimensional array, and that requires the second dimension the be defined in order to allocate a sequential block of memory. Instead, try one of the two approaches below:

• Multidimensional array:

  int p[][3] = {{1,2,3},{4,5,6}};
  

• Array of pointers to one dimensional arrays:

  int p1[] = {1,2,3};
  int p2[] = {4,5,6};
  int *p[] = {p1,p2};
  

The latter method has the advantage of allowing for sub-arrays of varying length. Whereas, the former method ensures that the memory is laid out contiguously.

Another approach that I highly recommend that you do NOT use is to encode the integers in string literals. This is a non-portable hack. Also, the data in string literals is supposed to be constant. Do your arrays need to be mutable?

int *p[] = (int *[]) {
    "\x01\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00",
    "\x04\x00\x00\x00\x05\x00\x00\x00\x06\x00\x00\x00"
};

That example might work on a 32-bit little-endian machine, but I'm typing this from an iPad and cannot verify it at the moment. Again, please don't use that; I feel dirty for even bringing it up.

The casting method you discovered also appears to work with a pointer to a pointer. That can be indexed like a multidimensional array as well.

int **p = (int *[]) { (int[]) {1,2,3}, (int[]) {4,5,6} };

First understand that "Arrays are not pointers".

int p[] = (int []) {1,2,3,4,5,6};

In the above case p is an array of integers. Copying the elements {1,2,3,4,5,6} to p. Typecasting is not necessary here and both the rvalue and lvalue types match which is an integer array and so no error.

int *p[] = (int *[]) {{1,2,3},{4,5,6}};

"Note I don't understand why I got a error in the first one,.."

In the above case, p an array of integer pointers. But the {{1,2,3},{4,5,6}} is a two dimensional array ( i.e., [][] ) and cannot be type casted to array of pointers. You need to initialize as -

int p[][3] = { {1,2,3},{4,5,6} };
  // ^^ First index of array is optional because with each column having 3 elements
  // it is obvious that array has two rows which compiler can figure out.

But why did this statement compile ?

char *p[] = {"one", "two"...};

String literals are different from integer literals. In this case also, p is an array of character pointers. When actually said "one", it can either be copied to an array or point to its location considering it as read only.

char cpy[] = "one" ;
cpy[0] = 't' ;  // Not a problem

char *readOnly = "one" ;
readOnly[0] = 't' ;  // Error because of copy of it is not made but pointing
                     // to a read only location.

With string literals, either of the above case is possible. So, that is the reason the statement compiled. But -

char *p[] = {"one", "two"...}; // All the string literals are stored in 
                               // read only locations and at each of the array index 
                               // stores the starting index of each string literal.

I don't want to say how big is the array to the compiler.

Dynamically allocating the memory using malloc is the solution.

Hope it helps !

Since nobody's said it: If you want to have a pointer-to-2D-array, you can (probably) do something like

int (*p)[][3] = &(int[][3]) {{1,2,3},{4,5,6}};

EDIT: Or you can have a pointer to its first element via

int (*p)[3] = (int[][3]) {{1,2,3},{4,5,6}};

The reason why your example doesn't work is because {{1,2,3},{4,5,6}} is not a valid initializer for type int*[] (because {1,2,3} is not a valid initializer for int*). Note that it is not an int[2][3] — it's simply an invalid expression.

The reason why it works for strings is because "one" is a valid initializer for char[] and char[N] (for some N>3). As an expression, it's approximately equivalent to (const char[]){'o','n','e','\0'} except the compiler doesn't complain too much when it loses constness.

And yes, there's a big difference between an initializer and an expression. I'm pretty sure char s[] = (char[]){3,2,1,0}; is a compile error in C99 (and possibly C++ pre-0x). There are loads of other things too, but T foo = ...; is variable initialization, not assignment, even though they look similar. (They are especially different in C++, since the assignment operator is not called.)

And the reason for the confusion with pointers:

• Type T[] is implicitly converted to type T* (a pointer to its first element) when necessary.
• T arg1[] in a function argument list actually means T * arg1. You cannot pass an array to a function for Various Reasons. It is not possible. If you try, you are actually passing a pointer-to-array. (You can, however, pass a struct containing a fixed-size array to a function.)
• They both can be dereferenced and subscripted with identical (I think) semantics.

EDIT: The observant might notice that my first example is roughly syntactically equivalent to int * p = &1;, which is invalid. This works in C99 because a compound literal inside a function "has automatic storage duration associated with the enclosing block" (ISO/IEC 9899:TC3).

The one that you are using is array of int pointers. You should use pointer to array :

int (*p)[] = (int *) {{1,2,3}, {4,5,6}}

Look at this answer for more details.

It seems you are confusing pointers and array. They're not the same thing! An array is the list itself, while a pointer is just an address. Then, with pointer arithmetic you can pretend pointers are array, and with the fact that the name of an array is a pointer to the first element everything sums up in a mess. ;)

int *p[] = (int *[]) {{1,2,3},{4,5,6}};      //I got a error

Here, p is an array of pointers, so you are trying to assign the elements whose addresses are 1, 2, 3 to the first array and 4, 5, 6 to the second array. The seg fault happens because you can't access those memory locations.

int p[][3] = {{1,2,3},{4,5,6}};              //it's okay

This is ok because this is an array of arrays, so this time 1, 2, 3, 4, 5 and 6 aren't addresses but the elements themselves.

char *p[] = (char *[]) {"one", "two"...};    // it's okay!

This is ok because the string literals ("one", "two", ...) aren't really strings but pointers to those strings, so you're assigning to p[1] the address of the string literal "one".
BTW, this is the same as doing char abc[]; abc = "abc";. This won't compile, because you can't assign a pointer to an array, while char *def; def = "def"; solves the problem.