C - Map one dimensional array into two - Why doesn't this work?

Question

First the error:

 error: a value of type "float *" cannot be used to initialize an entity of type "float (*)[2000]"

Heres the relevant code:

#define opType float
const int N = 2000;
opType *a_h,*b_h,*d_h, *a_d,*b_d, *d_d;
opType (*normalized_a)[N] = a_h;
opType (*normalized_b)[N] = b_h;
opType (*normalized_d)[N] = d_h;

opType (*normalized_a_d)[N] = a_d;
opType (*normalized_b_d)[N] = b_d;
opType (*normalized_d_d)[N] = d_d;

I am attempting to normalize it to two dimensions so that I can pass a double pointer to a function later on and reference the two dimensions using the shorthand bracket syntax. I am using CUDA so I must have one dimensional declarations for the copying of the memory. Is there any way to make the above work?

This construct is borrowed from http://forums.nvidia.com/index.php?showtopic=83941&view=findpost&p=475901

int linear[10*32]; // linear array
int (*twodim)[32] = linear; // interpreted as a two-dimensional [10][32] array.
assert(&twodim[2][4] == &linear[2*32 + 4]); // these are the same

Good catch, I'll keep that in mind for the future. – Joshua Enfield Dec 07 '10 at 22:25 — Joshua Enfield, Dec 07 '10 at 22:25

TToni · Accepted Answer · 2010-12-07T22:56:34.130

2

I don't see where the data for your array initialization comes from, but the error-message is quite clear: You declare 6 arrays, each of which contains 2000 float pointers. Of course you can't initialize one of these arrays with a single float pointer.

So in what form is your base data and how do you want to use it?

Edit: OK, based on your comment you have got something like

float myData[N][N];

filled with values. In this form you can already access it as myData[i][j]. Now, if you just cast it over to a float pointer like

float *myDataFlat = (float*)myData;

you can also access it through myDataFlat[i*N+j].

edited Dec 07 '10 at 22:56

answered Dec 07 '10 at 22:26

TToni

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See edit, my base data is allocated into N*N floats. I am basically trying to serialize `float x[N][N]`, but still be able to refer to it as if it were two dimensions (double pointer) in some form as well as in the serial form. – Joshua Enfield Dec 07 '10 at 22:32

score 0 · Answer 2 · edited May 23 '17 at 10:24

0

You can't initialise an array to a variable. You can initialise some members to a constant. In that case it would be:

opType (*normalized_a)[N] = {0}; /* all members 0 */

or you can use a loop or memset.

edited May 23 '17 at 10:24

Community

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answered Dec 07 '10 at 22:34

Dominique McDonnell

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score 0 · Answer 3 · answered Dec 07 '10 at 23:20

I am not an expert in C; i guess that the error is because of type mismatch (like it would have been in C++).
I think you have to do some casting on assignment. To make it readable you have to introduce a typedef:

typedef float opType;
const int N = 2000;
typedef opType (*TPointerToArray)[N]; // may be defined outside of a function
opType *a_h; // no value assigned yet
TPointerToArray normalized_a; // no value assigned yet

void MyCode()
{
    ...
    a_h = ...; // i don't know how the value is assigned; it should be done first
    ...
    normalized_a = (TPointerToArray)a_h; // "normalization"
    normalized_a[100][200] = 900; // usage
    ...
}

I assume you are using global variables just for fun; it doesn't matter much.
In this code, technically, normalized_a is a pointer-to-array but i can think about it as an array-of-arrays or two-dimensional-array.

C - Map one dimensional array into two - Why doesn't this work?

3 Answers3