How to cast simple pointer to a multidimensional-array of fixed size?

Question

I have a function that takes a pointer to a floating point array. Based on other conditions, I know that pointer is actually pointing to a 2x2 OR 3x3 matrix. (in fact the memory was initially allocated as such, e.g. float M[2][2] ) The important thing is I want to make this determination in the function body, not as the function argument.

void calcMatrix( int face, float * matrixReturnAsArray )
{
    // Here, I would much rather work in natural matrix notation
    if( is2x2 )
    {
        // ### cast matrixReturnAsArray to somethingAsMatrix[2][2]
        somethingAsMatrix[0][1] = 2.002;
        // etc..
    }
    else if(is3x3)
    { //etc...
    }

}

I am aware that I could use templates and other techniques to better address this problem. My question is really about how to make such a cast at the ### comment. Working in C++.

Hinterland

I think you're trying to solve the wrong problem (i.e. "how to cast?" is rarely the right problem). I wrote a simple solution to the problem of "how to use multi-dimensional arrays easily?" once: ideone.com/gytw7 – Luht

There is no way a float * points to a multidimensional anything (barring really terrible casting that you shouldn't be doing and I'd be surprised if a compiler let you). A float * points to a float, which might be the first value in a single-dimensional float array. But it does not point to any sub-array, as you would need for a multidimensional array. 2x2 and 3x3 are both 2D, so both could be a float **. Really, though, you'd be much better off creating (or finding) and using a dedicated Matrix class. – October

Ok, I could change my input argument to float **. But are you saying that in the case float aMDarray[3][3], the storage of elements is not guaranteed to be continuous? – Hinterland

@DragoonWraith: Sorry, but you are wrong. float a[2][2]; is not compatible with float**. – Rana

@BenVoigt: Yeah, you're right; they're not and I misspoke. I meant more that float * cannot be more than unidimensional and float ** is needed for two dimensions. – October

@DragoonWraith: That's still wrong, though. float a[2][2]; is a still a single sequence of float stored contiguously, but with compiler-provided two-dimensional address calculation. You can then write float* p = &a[0][0]; and do the index calculation yourself. – Rana

@BenVoigt: Ah, then I simply misunderstood entirely. My apologies. Thanks for the clarification, +1 on both. (Is it better to delete the comment, or not? Not sure of SO-tiquette here) – October

Related: Convert pointer to two dimensional array / Create a pointer to two-dimensional array – Jopa

Answer 1

float (*somethingAsMatrix)[2] = (float (*)[2]) matrixReturnAsArray;

Fabiolafabiolas

Thank you. This certainly works now. I'm just clarifying if I have a misconception about the guaranteed contiguous arrangement of elements in the declaration case float aMDarray[3][3]; – Hinterland

@NoahR: That's contiguous, and compatible with the usage you suggest. – Rana

I tried this technique and the Mac OS X C++ compiler said error: assigning to 'double (*)[LDN]' from incompatible type 'double (*)[LDN]'. Notice the two reported types are the same! So why are they not assignable? My code looks like this double (*F)[LDN]; F=(double (*)[LDN])AF; I also tried initializing in the declaration and got a similar error message. – Kolva

@Kolva perhaps your compiler is broken? If you can provide the full code that you're having trouble with then it'd be worth asking a separate question. – Fabiolafabiolas

Why does this work ? Is there a clear documentation about this cast and the types mixing pointers and arrays notation ? Thx – Smolt

@ZachB I am talking about the right term of the equality (matricReturnAsArray) not the left term which is clearly a declaration. – Smolt

@Smolt this is documented in the C Standard (ISO/IEC 9899:2011): 6.5.4 says that a cast-expression is a unary-expression preceded by any number of parenthesized type-names, and 6.7.7 says that a type-name is "syntactically a declaration for a function or an object of that type that omits the identifier". (So from float (*somethingAsMatrix)[2] we can remove the identifier somethingAsMatrix to get the type-name float (*)[2].) – Fabiolafabiolas

Thanks @Fabiolafabiolas that is very clear. Nevertheless my next question is : what do you call this type-name float (*identifier)[2] ? Is it a pointer to 2-sized-arrays of float ie a pointer to a space full of 2-sized-float-arrays ? – Smolt

@Smolt yes, that's right - cdecl.org says "declare identifier as pointer to array 2 of float". – Fabiolafabiolas

Answer 2

float * could point to the first element of an array of floats, and ought to be reinterpret_castable to that array type. And the result of that cast could point to the first element of a float [][] and so should be reinterpret_castable to that type, and so on. You ought to be able to compose such casts and just directly do

float (&arr)[2][2] = *reinterpret_cast<float (*)[2][2]>(matrixReturnAsArray);

An argument of the type float ** is not the same and should not be used this way.

To avoid undefined behavior the pointer must originate from an actual multi-dimensional array, and if the float* is used directly you cannot access more than the first row of the multi-dimensional matrix.

void foo(float *f) {
    f[3] = 10.;

    float (&arr)[2][2] = *reinterpret_cast<float (*)[2][2]>(f);
    arr[1][1] = 10.;
}

void main() {
    float a[2][2];
    foo(&a[0][0]); // f[3] = 10.; is undefined behavior, arr[1][1] = 10. is well defined

    float b[4];
    foo(&b[0]); // f[3] = 10.; is well-defined behavior, arr[1][1] = 10. is undefined
}

Given float arr[2][2]; nothing guarantees that &arr[0][1] + 1 is the same as &arr[1][0], as far as I have been able to determine. So although you can use a single dimensional array as a multi-dimensional array by doing f[i*width + j] you cannot treat a multi-dimensional array like a single dimensional array.

It's better to use C++'s compile-time type-safety instead of just relying on not accidentally passing the wrong thing or performing the wrong reinterpret_cast. To get type-safety using raw-arrays you should use references to the raw array type you want:

void foo(float (&f)[2][2]) {}
void foo(float (&f)[3][3]) {}

If you want to pass arrays by value you can't use raw arrays and should instead use something like std::array:

void foo(std::array<std::array<float,2>,2> f) {}
void foo(std::array<std::array<float,3>,3> f) {}

Dilettante

Isn't sizeof (T [N]) guaranteed to be exactly N * sizeof (T)? – Rana

I'm aware of a non-normative note in the standard that mentions that. If you can find anything normative please let me know. – Dilettante

5.3.3p2 directly states that it is required. Looks normative to me (not inside a "Note") – Rana

"When applied to an array, the result is the total number of bytes in the array. This implies that the size of an array of n elements is n times the size of an element." The issue is that I don't see anything that actually prohibits an array from containing extra space at the end. It's said to be 'implied' but I don't see how. – Dilettante

relevant: groups.google.com/forum/?fromgroups#!topic/comp.lang.c++/… – Rana

If nothing else, that line prohibits padding at the end of an array, because that line is normative. – Rana

If there's nothing else that actually implies what that line says is implied then I would not regard it as a normative statement, I would regard it only as a factually incorrect statement and a bug in the standard. – Dilettante

Answer 3

This sort of casting is always cleaner, and easier to deal with, with a judicious use of typedef:

typedef float Matrix_t[2][2];

Matrix_t* someThingAsMatrix = (Matrix_t*) matrixReturnAsArray;

If this is C++ and not C, though, you should create a matrix class. (Or better yet, look for an open source one.)

Yellowhammer

Use someThingAsMatrix[1][1] = 2.0f; gives: Incompatible types in assignment of 'float' to 'float [2]' – Hinterland

Typedef is still helpful, but you need: typedef float MatrixRow[2]; MatrixRow* someThingAsMatrix = (MatrixRow*) matrixReturnAsArray; which is equivalent to ecatmur's answer. – Rana

Answer 4

If I am right:

typedef float Matrix_t[2][2];  
Matrix_t &matrix = *(Matrix_t *)matrixReturnAsArray; 

or

float (&matrix2)[2][2] = *(float ( *)[2][2])matrixReturnAsArray;  

In C there is only the way with the pointer

Matrix_t *someThingAsMatrix = (Matrix_t *)matrixReturnAsArray;  

and access via:

(*someThingAsMatrix)[1][0] = ...  

Knuth