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I'm modifying existing code for a new project. I was instructed to remove dynamic memory allocation and use static declaration.

There is a variable arrp, earlier it was a double pointer, to which memory will be allocated using malloc and will be accessed as 2D array.

Now i have changed it as pointer to array i.e: char (*arrp)[];

The size of the 2D array to which arrp points to will be known only at runtime. My problem is if size is not declared compiler throws error('char (*)[]' : unknown size)

Please refer the following code, i did something like this

char (*arrp)[]; //This will be from different module, 
        //I have declared as local variable for our reference

char (*parr)[2];

char arr[3][2];

parr = &(arr[0]);   
arrp = (char (*)[])&(arr[0]);

//inside loops for i, j
...

printf("%c",parr[i][j]); // This works fine
printf("%c",arrp[i][j]); // Error :'char (*)[]' : unknown size) 

....
//Some code

It not possible to get the size of array when arrp is declared. Is there any way to eliminate this error?

share|improve this question
    
Sorry i forgot to mention, this project is in C –  Cks Apr 12 '12 at 10:27
    
What version of C? –  Christoffer Apr 12 '12 at 11:16
    
The application is built to work for various compilers like GCC 2.9.5, 3.4 (not sure about this), VC++ compiler etc –  Cks Apr 12 '12 at 13:09

2 Answers 2

up vote 0 down vote

Here's an article from Dr Dobbs that explains it better than I can hope to, and also includes hints for how to work with variable-length arrays in C90 and C99: http://www.drdobbs.com/184401444

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up vote 0 down vote

A pointer to an array helps in jumping over whole arrays at a time. ( ie with a single increment ) It does this through the knowledge of the column width of the array to be jumped. So without the knowledge of the column size, I am afraid, your pointer to an array will be of no use to you.

But if you have a modern compiler which supports variable length arrays ( C99 ), then its quite simple

int foo ( int m, int n )
{
    int a[m][n];
    int (*ptr)[n]=a;
    a[0][2] = 78;
    printf("%d", ptr[0][2]);

}
share|improve this answer
    
when i directly assign the address of 2D array double pointer it crashes during execution. char *****pparr; char arr[3][2]; pparr = (char*****)arr; when i try to print as printf("%c",pparr[i][j]); it crashes. (sorry for poor indentation, i dint know how to indent comments) :( –  Cks Apr 12 '12 at 11:01
    
Can you show how you are accessing the 2D array using your double pointer? –  xeek Apr 12 '12 at 11:02
    
@Cks I edited my answer. Double pointers cant work, again, for the same reason that they dont know array width ( column size ). So I am afraid you cant proceed without knowing the column size. –  xeek Apr 12 '12 at 11:18
    
@Cks. Why cant you know the size of the 2D array at compile time? –  xeek Apr 12 '12 at 11:19
    
This application process C files and generate some information that user requests. So it depends on the input file that user selects. The double pointer or pointer to array which i want to use will be declared in our library. our application creates some intermediate file that uses this pointer and allocates memory based on the matching pattern found in C file. So number of matching pattern varies based on input. –  Cks Apr 12 '12 at 13:03

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