c pointer array argument

Question

I have a function(in some library) whose signature is this:

extern LIB3DSAPI void lib3ds_mesh_calculate_face_normals(Lib3dsMesh *mesh, float (*face_normals)[3]);

what does it expect in second argument?

I tried this:

        float   *norm_verts[3];
        norm_verts=(float(*)[3])malloc(3*sizeof(float[3])*mesh->nfaces);
        lib3ds_mesh_calculate_face_normals(mesh, norm_faces);

on the second line, it says Expression must be modifiable value and the third line says argument of type float** is incompatible with parameter of type float(*)[3]

My intuition was that float* [3] is just 3 pointers but why the hell is the * wrapped in brackets?

Hint: in Linux there are nifty cdecl and uncdecl tools for encoding and decoding cryptic C definitions to plain english. — hyde
– hyde, Commented Oct 26, 2012 at 14:37
I was just wondering, if this function computes the normals of the mesh, wouldn't you need an array with the same size as the number of faces in the mesh ? — iabdalkader
– iabdalkader, Commented Oct 26, 2012 at 14:53
@mux:ok, Yeah sorry, lib3ds_mesh_calculate_face_normals calculates normals per face which would need an array of same size and there's another function to calculate per vertex normals which takes an array thrice as big. So I messed up while copying the code..:) — prongs
– prongs, Commented Oct 26, 2012 at 15:44

pmg · Accepted Answer · 2012-10-26 14:37:57Z

2

float (*face_normals)[3] // face_normals is a pointer (to an array of 3 floats)
float *norm_verts[3];    // norm_verts is an array of 3 pointers (to float)

Pointers are not arrays, arrays are not pointers. I suggest you read comp.lang.c FAQ, starting with section 6.

answered Oct 26, 2012 at 14:37

pmg

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Robᵩ · Accepted Answer · 2012-10-26 14:37:36Z

1

The * is wrapped in brackets to give it tighter binding. The 2nd arg to lib3ds_mesh_calculate_face_normals is read "face_normals is a pointer to an array of 3 float`.

Try:

float   (*norm_verts)[3];
norm_verts=(float(*)[3])malloc(sizeof(*norm_verts)*mesh->nfaces);
lib3ds_mesh_calculate_face_normals(mesh, norm_vets);

answered Oct 26, 2012 at 14:37

Robᵩ

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Lightness Races in Orbit · Accepted Answer · 2012-10-26 14:36:49Z

My intuition was that float* [3] is just 3 pointers

It is.

It's also not what the code says.

The function is asking for a pointer to an array of three floats. The parentheses make sure that it is parsed that way by "binding" the * to the name and not the type.

Lib3dsMesh *mesh    = getMeshFromSomewhere();
float norm_faces[3] = {};

lib3ds_mesh_calculate_face_normals(mesh, &norm_faces);

In this manner, the function lib3ds_mesh_calculate_face_normals knows that it is dealing with the original, actual array norm_faces and not some copy and not some name decayed to a pointer without dimension information.

This is the way to do "out" arguments with arrays without having to pass a float* and a separate length argument.

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