Get address of a multidimensional static array

Question

How can I get the address of a multidimensional static array? For example, this is my array

char array[2][10] = {"word1", "word2"};

Is it possible to get the address to make me reference this array using a pointer like this?

char** pointer;

I tried &array or directly pointer = (char**)array; but it crashes at startup.

char** pointer = array is sufficient. If it crashes, you have bug somewhere else. You don't even need a cast. — Gunther Piez
– Gunther Piez, Commented May 31, 2011 at 21:22
post your code. pointer=(char**)array; should have worked, if it's crashing - there's some other problem. — oh whatever
– oh whatever, Commented May 31, 2011 at 21:23
Uhm it's crashing because I'm referencing 2nd char* (word2) with "pointer[0]"... Instead using just "pointer" all chars until '\0' is printed thus I see "word1". But why is illegal? char** pointer should be an array of char*, so this expression "pointer[0]" shouldn't be illegal... — Exor
– Exor, Commented May 31, 2011 at 21:33
read section 6 of the c-faq: arrays are not pointers and pointers are not arrays. — pmg
– pmg, Commented May 31, 2011 at 21:36
Thanks I knew the difference... Unluckily I left C for too much time to Java, and these are the things you've not to worry about using Java. You think only "every reference is a pointer" :D — Exor
– Exor, Commented May 31, 2011 at 22:26

Omri Barel · Accepted Answer · 2011-05-31 21:42:30Z

5

char **pointer means that a pointer is pointing to a pointer. So *pointer is expected to be a pointer (e.g. 4-byte value which can be interpreted as an address).

This is not the case with your array: it is a contiguous area in memory (20 bytes).

So when you try to convert the array to char ** your application crashes. It's not possible to do this conversion, char ** must point at a pointer.

edited May 31, 2011 at 21:42

answered May 31, 2011 at 21:25

Omri Barel

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3 Comments

Omri Barel Over a year ago

And credit to "Expert C Programming" by Peter Van Der Linden who explained so many things I thought I already knew.

Exor Over a year ago

Right... "array" can be seen as a simple char* pointer treated as a "bounded" matrix by the compiler. Thus I will get in error assigning this array to an array of pointers. I should initialize every pointer by my self, doing like this? ` char arr[2][20] = {"word1","word2"}; char **pointer = new char*[2]; pointer[0] = arr[0]; pointer[1] = arr[1];`

Omri Barel Over a year ago

@Exor: Yes, that would work. But it has to be done "manually", there is no automatic conversion, because the types are incompatible.

Zmaster · Accepted Answer · 2011-05-31 21:52:57Z

2

"array" is the address of the array in memory but it is not a char**. While you can cast it, the application will crash if you try

printf("%s", pointer[1]);

because in your case is probably the same as

printf("%s", (char *)(0x00000031));

since pointer[1] means "the second 4 byte pointer (assuming x86) starting from 'array'". pointer[0] MAY not crash but won't show "word1" either for the same reason.

You probably want (this is hard to remeber so i had to check online, hope it is correct):

char (*pointer)[10] = array;

Which is a pointer to an array of 10 chars. And if you use pointer[1] it now means "the second 10 chars block starting from 'array'".

edited May 31, 2011 at 21:52

answered May 31, 2011 at 21:41

Zmaster

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