2D-Array manipulation through pointer parameter

Question

I'm creating a set of functions that takes a pointer to a 2D-Array and fills the array with some data

This is how I got it right now:

17 void m4identity(float *m[4][4]) {
18    *m = (float[4][4]) { { 1, 0, 0, 0 },
19                         { 0, 1, 0, 0 },
20                         { 0, 0, 1, 0 },
21                         { 0, 0, 0, 1 } };
22 }

But unfortunately I get a compiler error:

linalg.c:18:7: error: incompatible types when assigning to type ‘float *[4]’ from type ‘float (*)[4]’

Questions:

What is the difference between (*)[4] and *[4]?
Is there a better way to do this?

I initially tried to return a pointer to the array created inside the function but this threw another compiler error because it would be out of scope. I also want to avoid allocating space for the array from within the function as that would be hard to control.

user529758 · Accepted Answer · 2013-03-25 21:10:12Z

1

What is the difference between (*)[4] and *[4]?

The [] declaration specifier has higher precedence, so float *arr[4] declares an array of 4 pointers-to-float, while float (*arr)[4] declares a pointer-to-array-of-4-float.

Is there a better way to do this?

Just let the array decay into a pointer and use assignments:

void m4identity(float m[4][4])
{
    memset(m, 0, 4 * sizeof(m[0]));
    for (int i = 0; i < 4; i++) {
        m[i][i] = 1;
    }
}

answered Mar 25, 2013 at 21:10

user529758

Sign up to request clarification or add additional context in comments.

1 Comment

newacct Over a year ago

note that this is exactly the same as declaring void m4identity(float (*m)[4])

Collectives™ on Stack Overflow

2D-Array manipulation through pointer parameter

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related