I have declared an array char **arr; How to initialize the memory for the 2D char array.
16 Answers
One way is to do the following:
char **arr = (char**) calloc(num_elements, sizeof(char*));
for ( i = 0; i < num_elements; i++ )
{ arr[i] = (char*) calloc(num_elements_sub, sizeof(char));
}It's fairly clear what's happening here - firstly, you are initialising an array of pointers, then for each pointer in this array you are allocating an array of characters.
You could wrap this up in a function. You'll need to free() them too, after usage, like this:
for ( i = 0; i < num_elements; i++ )
{ free(arr[i]);
}
free(arr);I think this the easiest way to do things and matches what you need.
4There are two options for allocating an array of type char **
I've transcribed these 2 code samples from the comp.lang.c FAQ (which also contains a nice illustration of these two array types)
Option 1 - Do one allocation per row plus one for the row pointers.
char **array1 = malloc(nrows * sizeof(char *)); // Allocate row pointers
for(i = 0; i < nrows; i++) array1[i] = malloc(ncolumns * sizeof(char)); // Allocate each row separatelyOption 2 - Allocate all the elements together and allocate the row pointers:
char **array2 = malloc(nrows * sizeof(char *)); // Allocate the row pointers
array2[0] = malloc(nrows * ncolumns * sizeof(char)); // Allocate all the elements
for(i = 1; i < nrows; i++) array2[i] = array2[0] + i * ncolumns;You can also allocate only one memory block and use some arithmetic to get at element [i,j]. But then you'd use a char* not a char** and the code gets complicated. e.g. arr[3*ncolumns + 2] instead of arr[3][2]
You might be better off with a one dimensional array:
char *arr = calloc(WIDTH*HEIGHT, sizeof(arr[0]));
for (int y=0; y<HEIGHT; y++) for (int x=0; x<WIDTH; x++) arr[WIDTH*y+x] = 2*arr[WIDTH*y+x];
free(arr);Use the following trick:
typedef char char2D[1][1];
char2D *ptr;
ptr = malloc(rows * columns, sizeof(char));
for(i = 0; i < rows; i++) for(j = 0; j < columns; j++) (*ptr)[i][j] = char_value;char **array;
int row,column;
char temp='A';
printf("enter the row");
scanf("%d",&row);
printf("enter the column");
scanf("%d",&column);
array=(char **)malloc(row*sizeof(char *));
for (int i=0;i<row;i++)
{ array[i]=(char*)malloc(column*sizeof(char));
}By 2D char array, if you mean a matrix of strings then it may be done in the following way.
int nChars = 25; // assuming a max length of 25 chars per string
int nRows = 4;
int nCols = 6;
char *** arr = malloc(nRows * sizeof(char **));
int i;
int j;
for(i = 0; i < nCols; ++i)
{ arr[i] = malloc(nCols * sizeof(char *));
}
for(i = 0; i < nRows; ++i)
{ for(j = 0; j < nCols; ++j) { arr[i][j] = malloc(nChars * sizeof(char)); sprintf(arr[i][j], "Row %d Col %d", i, j); }
}To print the 2D char array(matrix of strings(char arrays))
for(i = 0; i < nRows; ++i)
{ for(j = 0; j < nCols; ++j) { printf("%s ", arr[i][j]); } printf("\n");
}Result is
Row 0 Col 0 Row 0 Col 1 Row 0 Col 2 Row 0 Col 3 Row 0 Col 4 Row 0 Col 5
Row 1 Col 0 Row 1 Col 1 Row 1 Col 2 Row 1 Col 3 Row 1 Col 4 Row 1 Col 5
Row 2 Col 0 Row 2 Col 1 Row 2 Col 2 Row 2 Col 3 Row 2 Col 4 Row 2 Col 5
Row 3 Col 0 Row 3 Col 1 Row 3 Col 2 Row 3 Col 3 Row 3 Col 4 Row 3 Col 5 
