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I got strucure

typedef struct 
{char *cells;}
 Map;

and cells suppose to be pointer to array of rows(in rows are integers on every position). I don't know how to access for example to number on 3. position in 2. row. I have stared with some array[3][3], but I don't know how to connect them with this struct. I tried 

Map nextmap;
nextmap.cells[0] = array[0][0];

But I got only first number, which is clear. How can I get to other positions? Thanks in advance. EDIT: renaming the structure .. .

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7
  • Note: You're dereferencing an indeterminate pointer, and thereby invoking undefined behaviour. Commented Sep 27, 2013 at 17:56
  • using new as the name for your structure is pretty iffy... Commented Sep 27, 2013 at 18:17
  • Why must you use char *cells; (as you indicated in a comment to an answer) instead of a different declaration for cells? Is the array to which cells will point an actual two-dimensional array (declared with something like char array[rows][columns];) or a pointer to pointers? Is the number of columns in the array fixed at compile time or stored in a variable? Commented Sep 27, 2013 at 19:08
  • because it's in a assigment in our school project :/ Commented Sep 27, 2013 at 19:10
  • @JakubFedor: Please answer the remaining two questions. Commented Sep 27, 2013 at 20:05

3 Answers 3

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When you did Map nextmap;, you created an uninitialized Map struct. When you did nextmap.cells[0] = array[0][0]; you dereferenced (i.e. followed) the uninitialized pointer, and stored a value at the random memory it points at.

If you want to initialize the cells structure, you can do something as simple as nextmap.cells = array[0]; That will cause nextmap.cells to point at array. Note that it's not copying the contents; just pointing at them. That means that if you change the values through cells, you'll be modifying the values in arrays.

(Also, using 'new' as a variable name is perfectly acceptable in C, but you're likely to confuse any C++ programmers reading your code, since 'new' is an operator in that language.)

new now changed to nextmap in question

Edited to correct the type mismatch in nextmap.cells assignment.

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5 Comments

new.cells = array will cause problems because the type of new.cells is not appropriate for a two-dimensional array.
Ye, I know, but I have to use one dimensional array to point to array of rows. EDIT: assigment from incompatible pointer type EDIT2: I have to use this struct
Either the declaration of cells should be changed to match the type of array or (less desired) a cast should be inserted into the assignment.
If you want to point to the row, you can do new.cells = array[0];, assuming that array is a char[][]. The type of array[0] would then be a char[], which automatically decays to char*.
After you've assigned arrays[0] to new.cells, the first item in the row will be new.cells[0]. The second item will be new.cells[1].
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Given an array char array[][NumberOfColumns] (the first dimension is irrelevant and is omitted here; it would be needed when the array is defined), you can set a pointer to the first element of the array with:

nextmap.cells = &array[0][0];

Then you can access an element in the array, array[i][j], by calculating its position within the array, with either of these two expressions:

*(nextmap.cells + i*NumberOfColumns + j)
nextmap.cells[i*NumberOfColumns + j]

Two-dimensional arrays generally ought to be addressed as two-dimensional arrays. Calculating the position manually is poor practice if done without good reason. If this school assignment did not have a good reason for this, then it is a bad assignment.

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0

First of all new is not a good name for a variable.

new now changed to nextmap in question

Second of all in your case cells should be a double pointer, like this

char ** cells;

Or a pointer to a 2D array, like

char (*cells)[N][N];

where N is a constant you want to use.

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1 Comment

char (*cells)[N][N]; is an inappropriate declaration if cells is to be assigned with cells = array;, where array is declared as char array[N][N];. If you keep the current declaration, cells would have to be assigned with cells = &array;, and elements would have to be accessed with three dereferences, as (*cells)[i][j].

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