Sorry for asking the already answered question, I am a newbie to C and don't understand the solutions. Here is my function
int rotateArr(int *arr) {
int D[4][4];
int i = 0, n =0;
for(i; i < M; i ++ ){
for(n; n < N; n++){
D[i][n] = arr[n][M - i + 1];
}
}
return D;
}
It throws an error
main.c|23|error: subscripted value is neither array nor pointer nor vector|
on line
D[i][n] = arr[n][M - i + 1];
What's wrong? I am just setting the value of an array element to another array element.
The arr passed is declared as
int S[4][4] = { { 1, 4, 10, 3 }, { 0, 6, 3, 8 }, { 7, 10 ,8, 5 }, { 9, 5, 11, 2} };
The problem is that arr
is not (declared as) a 2D array, and you are treating it as if it were 2D.
You have "int* arr" so "arr[n]" is an int, right? Then your "[M - 1 + 1]" bit is trying to use that int as an array/pointer/vector.
You are not passing your 2D array correctly. This should work for you
int rotateArr(int *arr[])
or
int rotateArr(int **arr)
or
int rotateArr(int arr[][N])
Rather than returning the array pass the target array as argument. See John Bode's answer.
the second subscript operator is invalid here. You passed a int * pointer into function, which is a 1-d array. So only one subscript operator can be used on it.
Solution : you can pass int ** pointer into funciton
Except when it is the operand of the sizeof
or unary &
operator, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T
" is converted ("decays") to an expression of type "pointer to T
", and the value of the expression is the address of the first element of the array.
If the declaration of the array being passed is
int S[4][4] = {...};
then when you write
rotateArr( S );
the expression S
has type "4-element array of 4-element array of int
"; since S
is not the operand of the sizeof
or unary &
operators, it will be converted to an expression of type "pointer to 4-element array of int
", or int (*)[4]
, and this pointer value is what actually gets passed to rotateArr
. So your function prototype needs to be one of the following:
T rotateArr( int (*arr)[4] )
or
T rotateArr( int arr[][4] )
or even
T rotateArr( int arr[4][4] )
In the context of a function parameter list, declarations of the form T a[N]
and T a[]
are interpreted as T *a
; all three declare a
as a pointer to T
.
You're probably wondering why I changed the return type from int
to T
. As written, you're trying to return a value of type "4-element array of 4-element array of int
"; unfortunately, you can't do that. C functions cannot return array types, nor can you assign array types. IOW, you can't write something like:
int a[N], b[N];
...
b = a; // not allowed
a = f(); // not allowed either
Functions can return pointers to arrays, but that's not what you want here. D
will cease to exist once the function returns, so any pointer you return will be invalid.
If you want to assign the results of the rotated array to a different array, then you'll have to pass the target array as a parameter to the function:
void rotateArr( int (*dst)[4], int (*src)[4] )
{
...
dst[i][n] = src[n][M - i + 1];
...
}
And call it as
int S[4][4] = {...};
int D[4][4];
rotateArr( D, S );
Source: Stackoverflow.com