Changing a pointer to an array in C

Question

I have got a structure and in it a pointer to a 2D array. But when I try to assign an actual 2D array to that pointer I do not succeed - compiler says that my pointer is a pointer to a 1D array.

Here's my code:

typedef GLfloat Vertex2f[2];
typedef GLfloat TextureCoordinate[2];

typedef struct { 
    GLuint texture_name;  // OpenGL ID of texture used for this sprite 
    Vertex2f *vertices; // array of vertices 
    TextureCoordinate *texture_coords; // texture vertices to match

    GLubyte *vertex_indices;
} game_sprite;

void loadState()
{
    game_sprite ballSprite;

    createAndLoadTexture("ball.png", &ballSprite.texture_name);

    const Vertex2f tempVerticesArray[4] =  {
        {-100.0f, -100.0f},
        {-100.0f, 100.0f},
        {100.0f, 100.0f},
        {100.0f, -100.0f}
    };

    ballSprite.vertices = &tempVerticesArray; //The problem appears to be here
}

How can I make it work?

Thanks.

Adam Rosenfield · Accepted Answer · 2010-01-01 15:16:49Z

You have two problems. First is that tempVerticesArray is const. You can't assign a pointer to a const value (&tempVerticesArray) to a pointer to a non-const variable (ballSprite.vertices) without a typecast, so the compiler is complaining. You should modify the vertices data member to be of the type const Vertex2f *, assuming you're not actually modifying that data ever.

The second problem is that as soon as loadState() ends, the variable tempVerticesArray goes out of scope, and so any dangling pointers to it (specifically ballSprite.vertices) are invalid. You should make tempVerticesArray a static variable so it's not a stack variable that can go out of scope. This is assuming that the ballSprite object is used after that function ends, which I'm guessing it does based on context.

If you do need to modify your vertices after initialization, you'll need to allocate for each ballSprite its own set of vertex data (e.g. using malloc()) and copy the vertex data in (e.g. using memcpy()). If you don't, all ballSprite instances will share a pointer to the same vertex data, and when you modify it, they will all be affected.

Mitch Wheat · Accepted Answer · 2010-01-01 15:03:09Z

3

My C is a little rusty, but shouldn't that be:

ballSprite.vertices = tempVerticesArray;

answered Jan 1, 2010 at 15:03

Mitch Wheat

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4 Comments

Richard Pennington Over a year ago

In C/C++ tempVerticesArray has type "Vertex2f*" and &tempVerticesArray has type "Vertex2f[2]*". The first is a pointer to the (first) element, the second is a pointer to an array.

Adam Rosenfield Over a year ago

No. Both ballSprite.vertices = tempVerticesArray and ballSprite.vertices = &tempVerticesArray are equivalent. tempVerticesArray has type const Vertex2f[2], and it decays into a pointer to a first element (of type const Vertex2f*). &tempVerticesArray has type const Vertex2f (*)[2], which is a compatible type with const Vertex2f*. The problem is with const correctness.

Richard Pennington Over a year ago

No, they are not compatible pointer types. One is a pointer to a Vertex2f, the other is a pointer to an array of Vertex2f's.

John Bode Over a year ago

The types of the expressions are not compatible; remember that Vertext2F is a typedef for GLFloat [2], so the expression ballSprite.vertices is of type GLFloat (*)[2] and the type of the expression &tempVerticesArray is GLFloat (*)[4][2], which are not compatible; the expression tempVerticesArray has type GLFloat [4][2], which will be implicitly converted to GLFloat (*)[2]. EDIT The const issue doesn't help.

Alon · Accepted Answer · 2010-01-02 18:20:02Z

3

in C an array is in fact a pointer to the first member, so when you do

ballSprite.vertices = &tempVerticesArray;

you are actually trying to assign to ballSprite.vertices the value of a pointer to a pointer

drop the & and you'll be fine with ballSprite.vertices = tempVerticesArray;

edited Jan 2, 2010 at 18:20

answered Jan 1, 2010 at 15:10

Alon

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3 Comments

Georg Fritzsche Over a year ago

In C++, an array is not a pointer, but decays to one in certain contexts.

jamesdlin Over a year ago

@gf: That's true in C as well.

Georg Fritzsche Over a year ago

@jamesdlin: ah, wasn't entirely sure.

phtrivier · Accepted Answer · 2010-01-01 15:11:39Z

2

Just to clarify, I don't think

Vertex2f *vertices

declares a "2D array", it is a 1-dimensionnal array of Vertex2f.

But as pointed by Mitch, the issue is probably that tempVerticesArray, as declared here, is a 1d-array, which is the same as a pointer to a Vertex2f. So &tempVerticesArray is the address of this pointer (a Vertex2f * * ).

answered Jan 1, 2010 at 15:11

phtrivier

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Comments

rmn · Accepted Answer · 2010-01-01 15:12:33Z

1

You could use either

ballSprite.vertices = tempVerticesArray;

or

ballSprite.vertices = &tempVerticesArray[0];

answered Jan 1, 2010 at 15:12

rmn

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Comments

GrkEngineer · Accepted Answer · 2010-01-01 15:18:28Z

1

You might want to allocate memory for the pointer to hold the data. Then memcpy the temporary array to the newly allocated memory. If you're just trying to make the pointer reference the array directly, you might have an issue when you leave that function. The temporary array made on the stack will be freed before you can use the pointer elsewhere.

answered Jan 1, 2010 at 15:18

GrkEngineer

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