Many dimension array

ratsietsi · 09-17-2006, 02:10 AM

I am trying to declare a two dimensional array without giving it any size. The program compiles without any errors but i keep on getting a segmentation error and I can't see where it comes from. This is how the code looks like:

----from the class------------------
private:
int **LDM;
============================================================

----this part is from a different segment of the code-----

LDM = new **int;

for(i = 0; i <= 4; i++){
for( j = 0; j < k; j++)
{
temp = patern1[i] - patern2[j];
if(temp < 0)
temp = -1 * temp;
LDM[i][j] = temp;
cout << "Local Distance Matrix: " << LDM[i][j] << endl;
}

zhangmaike · 09-17-2006, 05:06 AM

Quote:

I can't see where it comes from.

Using a debugger (such as gdb) would help with this.

Quote:

I am trying to declare a two dimensional array without giving it any size.

Well, there's your problem. This is not the way things are done in C or C++.

C and C++ will not do anything that you don't explicitly tell them to do. If you don't say "allocate memory for my two dimensional array of size X by Y", they won't... and you haven't. Besides, even if C++ were designed to do this sort of allocation for you, the math necessary to translate a two-dimensional coordinate into a one-dimensional memory address requires the size of at least one of the dimensions to be known, and you aren't specifying that either.

Now, look at the line in your code which you intend to allocate an array with:

Code:

LDM = new **int;

You are not creating a new int array, you are creating a new **int: a pointer to a pointer to an int. On a typical 32-bit machine, you've just allocated only 4 bytes of memory.

This is the source of your segmentation fault, since you eventually try to access memory that hasn't been allocated (the usual meaning of a segmentation fault).

You could accomplish what you're trying to do with a vector of vectors, or by using the new operator properly (the method I'd recommend, since the dimensions of your array are apparently known before you enter the nested for-loops).

I suggest finding a good C++ tutorial through google, or check out cppreference.com if you need a quick reference to some of the STL classes like vector (I've got it bookmarked). There are also plenty of good C++ books out there that could teach you these things.

Flesym · 09-17-2006, 05:38 AM

Also:

Code:

int **LDM;
LDM = new **int;

Beside the fact, that this does not, what you want it to do (as 'zhangmaike' said), I can't believe, that you are able to compile this. Not only that it is a wrong syntax of new, semantically you try to assign a "int***" to a "int**"?? -I've no possibilietes to test this at the moment, so maybe I'm wrong and there exists some strange hacks, but I would code:

Code:

int **LDM;
LDM = new int*;

... But as mentioned, this isn't what you want at all.

zhangmaike · 09-17-2006, 01:28 PM

I've just tested the following code for whether or not it compiles:

Code:

int main() {
        int **LDM;
        LDM = new **int;

        delete LDM;
        return 0;
}

It does not. It complains of a syntax error before the * token on line 3. Moving th ** to the other side of int causes this error to become:

Code:

error: cannot convert `int***' to `int**' in assignment

Flesym is correct.

ratsietsi: I can only assume that you are using a non-standard C++ compiler, though it would surprise me greatly if there exists such a C++ compiler which would compile the above code (or yours) without errors.