Cannot convert ‘float****’ to ‘float (*)[128][3][4]’ in initialization [duplicate]

Question

I have the following 2 typedef's:

typedef float matrix_3x4[3][4];
typedef matrix_3x4 bone_mat[128];

And the following function definition:

float ****get_bones();

Why is the following code giving me a type conversion error?

bone_mat *mat = get_bones();

Error:

error: cannot convert ‘float ****’ to ‘float (*)[128][3][4]’ in initialization
   85 |     bone_mat *mat = get_bones();
      |                     ~~~~~~~~~^~
      |                              |
      |                              float****

Tried changing the type of mat to bone_mat or bone_mat**, but shouldn't float(*)[128][3][4] be the same as float****?

Casting the return value of the function to (bone_mat*) doesn't work either.

Not an answer, but once you get beyond more than 2 levels of indirection, consider changing your types to nested structures and arrays thereof. It's hard to reason about a float ****. — pmacfarlane
– pmacfarlane, Commented Jul 28, 2023 at 20:42
No. Because dimensions of an array object are a bound its type. The float**** has no dimensions information in it thus it is not a pointer to array even though it admits ptr[a][b][c][d] syntax — tstanisl
– tstanisl, Commented Jul 28, 2023 at 20:55
@trxgnyp1 Do you have access to the source code of get_bones()? If so, can you post it? — Andrew Henle
– Andrew Henle, Commented Jul 28, 2023 at 21:12

trxgnyp1 · Accepted Answer · 2023-07-30 22:14:07Z

2

A pointer is not an array. A pointer can point to an element of an array. float **** is a pointer to a pointer to a pointer to a pointer to float, a very unlikely object to come across in a real life project.

mat is a pointer to a bone_mat which itself is an array or 128 arrays of 3 arrays of 4 floats, a very different type of beast.

Should you post the source code to get_bones(), I would be surprised if it returns an actual float ****, I would bet it allocates memory for a bone_mat and converts the object pointer to a type inconsistent with the allocation size and block usage.

Quoting Brian61354270:

Think about the memory and addresses involved. float*** * is a pointer to a float***. That means that when you dereference a float****, you arrive at a location in memory that contains a float***. Not a float, but a pointer to another pointer. Dereference that pointer, and you arrive to a location that contains a float**, Still not a float, but another pointer to a pointer. I could keep going. Contrast that with float (*)[128][3][4], which is a pointer to a chunk of memory with 128 * 3 * 4 contiguous floats.

edited Jul 30, 2023 at 22:14

trxgnyp1

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answered Jul 28, 2023 at 20:59

chqrlie

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1 Comment

trxgnyp1 Over a year ago

I do not have the source code of the function, only the definition. I am marking the answer as valid, but Brian61354270's comment explained way better.

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