2D char array and pointer

Is this okay:

Code:

#include <stdio.h>

void testfunc(char **ray) {
	printf("%c\n",ray[2][2]);		
}


int main () {
	char *ray[5][12]={"this","that","and","more"};
	testfunc((char**)ray);
	return 0;
}

I have been doing it this way for awhile, since it's the only way to pass a "list" to a parameter "char **" without using malloc, and there is no way to assign a functional pointer to char list[5][12]. But today I am thinking that char *list[5][12] might not be what I hope that it is. Altho if it's not, I'm not sure what it would be...

No, because a char ** is not the same as a char [5][12] or some such. They may appear similar, but when you pass the pointer to char[5][12], you are actually passing a pointer to (*char)[12], not the address of an array of pointers to char.

In this case, you must supply all but the left-most size, e.g.

Code:

void testfunc(char ray[][12]) {
	printf("%c\n",ray[2][2]);		
}

--
Mats

Further, this is an array of const char* pointers, not a 2D array, which makes it even more wrong.
Of course, since it's an array of pointers, it just happens to work:

Code:

#include <stdio.h>

void testfunc(const char** ray)
{
	printf("%c\n", ray[2][2]);		
}

int main ()
{
	const char* ray[] = { "this", "that", "and", "more" };
	testfunc(ray);
	return 0;
}

Originally Posted by Elysia

Further, this is an array of const char* pointers, not a 2D array, which makes it even more wrong.

The code posted is valid (besides the passing to the function part, that is), it initializes a set of 12-element arrays to the values given by the initializer list.

--
Mats

Originally Posted by matsp

No, because a char ** is not the same as a char [5][12] or some such. They may appear similar, but when you pass the pointer to char[5][12], you are actually passing a pointer to (*char)[12], not the address of an array of pointers to char.

By (char*)[12], do you mean:

• an array of twelve pointers
• a pointer with the size of 12?

If it is the former, then there really is not a problem (since all the elements will have the same "number", each element of (char*)[12] will be an element of *ray[5]. I suppose the stack size of **ray in testfunc() will be 12x instead of 5x.

What I am mostly worried about is the notation *ray[5][12] implies I could actually have a 3Dish array (a 2D pointer array) and that the "elements" to which I am assigning strings may have sizeof(char*) and not 12. For example, I can now compile this with char *ray[5][12]:

Code:

printf("%s\n",ray[0][0][0]);

where char ray[5][12] would produce a subscipted value error. It segfaults tho.

SInce I don't seem to be producing an overflow (no GUI debugger...) with it, apparently *ray[5][12] is really a 2D char array (a "list") with a pointer to it? Just using ray[5][12] makes a **pointer impossible.

To make a long story short: you don't see a potential overflow in main(), just a possible exceeding of boundaries in testfunc() with the number of elements?

I'm guessing this is because it's easier for the compiler to use an array of pointers with 12 bytes assigned to each rather than to use a pointer to an array of elements with a size of 12 each. So there is no pointer to such a thing possible.

ps. I really really want **char as the argument to testfunct()

Originally Posted by matsp

The code posted is valid (besides the passing to the function part, that is), it initializes a set of 12-element arrays to the values given by the initializer list.

--
Mats

Perhaps the code is correct, but I do question if it's intended. Clearly, it is an array of strings, and as thus, in its current form, should be an array of pointer or a 2D array.

Originally Posted by MK27

By (char*)[12], do you mean:

• an array of twelve pointers
• a pointer with the size of 12?

No, it's a pointer to an array of 12 elements of type char*.
The actual type should be char (*)[5].

Code:

#include <stdio.h>

void testfunc(char (*ray)[5])
{
	printf("%c\n", ray[2][2]);		
}

int main ()
{
	char ray[12][5] = {"this", "that", "and", "more"};
	testfunc(ray);
	return 0;
}

Sorry, I missed the (very obvious) red star in the original post.

Still, a 2D array is not a pointer to pointer. If you want to do that, then do what Elysia says.

If you want a 2D array of char, and need a pointer to that [e.g. you want to modify the content of the strings in an array of strings], then you need to pass the length of the strings.

--
Mats

Originally Posted by Elysia

Of course, since it's an array of pointers, it just happens to work:

That's it. I just noticed it functions as a 3D array in main.

I imagine yours does, because array are laid out sequentially in memory. It just happens to work, but I would definitely say it's undefined.
EDIT: It might just work because you have a pointer to the array (all arrays, 2D or whatever are essentially just one pointer anyway), and as the array is storing pointers, you have an extra pointer in the type and it... works. But if you change it to a 2D array or 3D array you will get trouble.

Hmmm...okay I just noticed this seems to work very nicely, it's only slightly more limited but it still saves a lot of potential malloc and strcpy. I suppose it is the same as just using char[][12] except I can use actual pointers in testfunc() (rather than a local copy):

Code:

#include <stdio.h>
#include <string.h>

typedef struct {
	char X[256];
} string;

void testfunc(string *ray) {
	printf("%s\n",(char*)&ray[2]);
	strcpy((char*)&ray[3],"well done");
}


int main () {
	string ray[4]={"this and that","more and","hello world"};
	testfunc(ray);
	printf("%s\n",(char*)&ray[3]);		
	return 0;
}

You actually don't even need the (char*) casts. The only issue seems to be that -Wall claims:

Code:

test2.c: In function ‘main’:
test2.c:14: warning: missing braces around initializer
test2.c:14: warning: (near initialization for ‘ray[0]’)

Is this significant and can I get out of it?

I believe it must be:

Code:

string ray[4] = { { "this and that" }, etc };

One initializer for each struct in the array and one for each member in the struct.

You can get rid of the warning by doing:

Code:

	string ray[4]={{"this and that"},{"more and"},{"hello world"}};

Which is exactly what the compiler told you anyways.

--
Mats

Thanks again people.

Perhaps you meant to print the string instead of the character it points to - and here's my 2c.

Code:

#include <stdio.h>

void testfunc(const char *ray[][12])
{
    printf("%s\n",ray[0][2]);
}

int main()
{
    const char *ray[5][12]={"this","that","and","more"};
    testfunc(ray);
    return 0;
}

Originally Posted by itCbitC

Perhaps you meant to print the string instead of the character it points to - and here's my 2c.

Perhaps you should have -- nevermind. Anyway all the code I posted works in the way I describe it as working, I wasn't asking for that kind of fix.