As a non-C/C++ expert I always considered square brackets and pointers arrays as equal.
ie :
char *my_array_star;
char my_array_square[];
But I noticed that when use in a structure/class they don't behave the same :
typedef struct {
char whatever;
char *my_array_star;
} my_struct_star;
typedef struct {
char whatever;
char my_array_square[];
} my_struct_square;
The line below displays 16, whatever
takes 1 byte, my_array_pointer
takes 8 bytes.
Due to the padding the total structure size is 16.
printf("my_struct_star: %li\n",sizeof(my_struct_star));
The line below displays 1, whatever
takes 1 byte, my_array_pointer
isn't taken in account.
printf("my_struct_square: %li\n",sizeof(my_struct_square));
By playing around I noticed that square brackets are used as extra space in the structure
my_struct_square *i=malloc(2);
i->whatever='A';
i->my_array_square[0]='B';
the line blow displays A:
printf("i[0]=%c\n",((char*)i)[0]);
the line blow displays B:
printf("i[1]=%c\n",((char*)i)[1]);
So I cannot say anymore that square brackets are equals to pointers. But I'd like to understand the reason of that behavior. I'm afraid of missing a key concept of that languages.
Array in C is used to store elements of same types whereas Pointers are address varibles which stores the address of a variable. Now array variable is also having a address which can be pointed by a pointer and array can be navigated using pointer.
An array is a pointer, and you can store that pointer into any pointer variable of the correct type.
Square brackets are used to index (access) elements in arrays and also Strings. Specifically lost[i] will evaluate to the ith item in the array named lost.
An array is represented by a variable that is associated with the address of its first storage location. A pointer is also the address of a storage location with a defined type, so D permits the use of the array [ ] index notation with both pointer variables and array variables.
Arrays and pointers don't behave the same because they're not the same at all, it just seems that way.
Arrays are a group of contiguous items while a pointer is ... well ... a pointer to a single item.
That single item being pointed to may well be the first in an array so that you can access the others as well, but the pointer itself neither knows nor cares about that.
The reason that arrays and pointers often seem to be identical is that, in many cases, an array will decay to a pointer to the first element of that array.
One of the places this happens is in function calls. When you pass an array to a function, it decays into a pointer. That's why things like the size of an array don't pass through to the function explicitly. By that I mean:
#include <stdio.h>
static void fn (char plugh[]) {
printf ("size = %d\n", sizeof(plugh)); // will give char* size (4 for me).
}
int main (void) {
char xyzzy[10];
printf ("size = %d\n", sizeof(xyzzy)); // will give 10.
fn (xyzzy);
return 0;
}
The other thing you'll find is that, while you can plugh++
and plugh--
to your hearts content (as long as you don't dereference outside of the array), you can't do that with the array xyzzy
.
In your two structures, there's a major difference. In the pointer version, you have a fixed size pointer inside the structure, which will point to an item outside of the structure.
That's why it takes up space - your 8-byte pointer is aligned to an 8-byte boundary as follows:
+----------------+
| 1 char variable|
+----------------+
| 7 char padding |
+----------------+
| 8 char pointer |
+----------------+
With the "unbounded" array, you have it inside the structure and you can make it as big as you want - you just have to allocate enough memory when you create the variable. By default (ie, according to the sizeof
), the size is zero:
+----------------+
| 1 char variable|
+----------------+
| 0 char array |
+----------------+
But you can allocate more space, for example:
typedef struct {
char whatever;
char my_array_square[];
} my_struct_square;
my_struct_square twisty = malloc (sizeof (my_struct_square) + 10);
gives you a variable twisty
which has a whatever
character and an array of ten characters called my_array_square
.
These unbounded arrays can only appear at the end of a structure and there can be only one (otherwise the compiler would have no idea where these variable length section began and ended) and they're specifically to allow arbitrarily sized arrays at the end of structures.
The my_array_square
member is what is called a "flexible" array member. Such arrays without a specified size can only appear at the end of a struct, and they don't contribute to its size. The intent is to manually allocate the rest of the space for as much elements as you need. Otherwise, the size of the array is determined at compile-time.
The usage pattern of such a struct would be as follows:
my_struct_square *s = malloc(sizeof(my_struct_square) + 5 * sizeof(char));
...
s->my_array_square[4]; // the last element of the array
In all other cases, the size of an array must be known at compile-time. Even the type of an array goes together with its size, i.e., int a[20]
is of type int[20]
, not just int[]
.
Also, understanding the difference between arrays and pointers is crucial. @paxdiablo has covered that quite well.
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