array definition error: why do I have this error?-CodePudding

EDIT: (SOLVED) I've rewritten comments that solved the problem:

"Pointers are not arrays. In case you actually had an array, which you don't, you can only initialize arrays during declaration, not assign to them in run-time".
"This is invalid C. You need A.data[0] = 1; A.data[1] = 42; etc."

and, of course, there's the accepted answer too.

(disclaimer: just ignore mat_constr function because it's not related to this question (it's mat_constructor of another question I've posted, because I'm trying to edit the mat_transpose function on my own); mat_constr creates the matrix by taking the address of the matrix variable, rows, and cols).

    void mat_constr(struct matrix* m, size_t rows, size_t cols) {
        m->rows = rows; m->cols = cols; 
        m->data = calloc(rows * cols, sizeof(double)); 
    }
// and, of course, mat_destroy to free the allocated memory. 
    void mat_destroy(struct matrix* m) {
        free(m->data); 
    }

this is a very simple exercise, yet I can't figure out how to define an array (correctly).

I have a

struct matrix {
size_t rows; 
size_t cols; 
double* data 
};

in the main function, I have to define a variable of type "struct matrix", and then I have to define the array. As follows:

int main(void) {
struct matrix A; 
    mat_constr(&A, 4, 4); 
    A.data = { /* a number */, /* another number* /, /* etc. */ }; 
}

the problem is that the compiler highlighted the first "{". in the array definition, and it says: "expected an expression". this is strange, because this is the definition style of the array, am I wrong?

CodePudding user response：

Assuming that mat_constr allocates memory as a single array of double, and makes the data member point to it, then you can copy to this array.

For example using memcpy and compound literals:

memcpy(A.data, (double[16]){ 1, 2, 3, ... }, 16 * sizeof(double));

_{[I made an assumption that mat_constr allocates a single array of 4 * 4 elements in my example above. Please don't let us make assumptions, create a proper Minimal, Reproducible Example instead.]}

You can of course copy a value to each element of the array individually instead:

A.data[0] = 1;
A.data[1] = 2;
A.data[2] = 3;
...

And if you want to set all elements to a single value you can of course use a loop:

for (unsigned i = 0; i < 16;   i)
{
    A.data[i] = 5;
}

Or if you want to set all elements to zero then either allocate with calloc:

A->data = calloc(A->rows * A->cols, sizeof(double));

Or use memset:

memset(A.data, 0, 16 * sizeof(double));

CodePudding user response：

The problem is that C has no notion of array litteral. And the equal sign can be used for 2 very different operations, assignment and initialization.

The brace enclosed expression can be used to initialize an array, but cannot be used to initialize a pointer nor can it be assigned to a pointer:

double data[] = {1., 2., 3.};  // array initialization: valid
double *data = {1.,  2.,  3.}; // INVALID attempt to initialize a pointer
double *d2;
d2 = {1., 2., 3.};             // INVALID assignment to a pointer