I want to make a pointer to an array of string(matrix[10][10]), but I get "initialization from incompatible pointer type" problem so how to fix it? And what is the problem?
void rotate(char matrix[10][10]){
char * pMatrix = matrix;
for(int j = 0; j < 10; j ++){
for(int i = 9; i >= 0; i --){
*pMatrix = matrix[i][j];
pMatrix ++;
}
}
}
btw, my task is to perform 90 degree clockwise rotation of matrix, but my algorithm is terrible I know
matrix is of type char (*)[10] (pointer to array of 10 char).
pMatrix is of type char * (pointer to char).
There is a type mismatch when you use:
char *pMatrix = matrix;
You try to assign a pointer to an array of 10 char to a pointer to char.
This is why you get the warning:
warning: initialization of '
char *' from incompatible pointer type 'char (*)[10]'
You need to dereference matrix
char *pMatrix = *matrix;
to get a pointer to char.
Could you say why my version (
char * pMatrix = matrix;) works for 1D array, but stops working for 2d array?
matrix' type is different when declared as char matrix[10] and parameter of a function. Then matrix is actual equivalent to char *. The assignment from char * to char * is correct.
Take a look at:
Difference between passing array and array pointer into function in C
C pointer notation compared to array notation: When passing to function
Note that if you provide an amount of elements like in your case doesn't matter. char matrix[10] is equal to char matrix[] which is furthermore equal to char *matrix.
No guarantee if your algorithm works beside that. If you got problems with that, please ask a different question.
There are so many ways but I prefer this one:
void rotate(char matrix[10][10]){
char *pMatrix = matrix[0];
for(int j = 0; j < 10; j ++){
for(int i = 9; i >= 0; i --){
*pMatrix = matrix[i][j];
pMatrix ++;
}
}
}
Or you could use this for better understanding
void rotate(char matrix[10][10]){
char * pMatrix = &matrix[0][0];
for(int j = 0; j < 10; j ++){
for(int i = 9; i >= 0; i --){
*pMatrix = matrix[i][j];
pMatrix ++;
}
}
}
Check this,
void rotate(char matrix[][10]){
char * pMatrix = &matrix[0][0];
for(int j = 0; j < 10; j ++){
for(int i = 9; i >= 0; i --){
*pMatrix++ = matrix[i][j];
}
}
}
One dimensional array:
Consider an array a of 4 integers as a[4].
Basic rule is both a and &a will point to same location.But they aren't pointing to the same type.
(1) &a will point to the entire array which is an int[]. The pointer type is int(*)[]
(2) a, when it decays to a pointer, will point to the first element of the array which is an int. The pointer type is int * Now consider two-dimensional array.
Two dimensional array
Consider an array containing two 1D arrays, with each of them having two elements; a[2][2].
Since the number of dimensions increases, we have one more level of hierarchy i.e. &a, a and *a will point to the same location, but they aren't pointing to the same type.
(1) &a will point to the entire array which is int[][]. The pointer type is int(*)[][].
(2) a, when it decays to a pointer, will point to the first element of the 2D array which is an int[]. The pointer type is int(*)[]
(3) By using *a, we are de-referencing the pointer to a 1D array. Hence we will have a int * pointing to the first integer value of the 2D array.
Now considering your case,
(1) In your case is like the first case where you are just using 1 D array, and the one which I suggested as &matrix[0][0], the pointer points to the entire array. Hope this is clear now.
(2) Also, you need not pass the first dimension in the function.
char *** pointer_to_array_of_strings;