Adding matrix as pointer

Question

I have a function for what I pass a matrix and a variable as a pointer (I simplified it, maybe in this form it doesn't make sense, but I didn't want to paste the whole code :)

uint64_t returndata = 0;
uint32_t binary_lock = 0;

void initLamp(uint8_t time, const uint8_t *datatable, uint64_t *outdata)
{
    for(uint8_t k = 0; k < ROW; k++)
    {
        if(*(datatable + k * COLS) == time)
        {
            binary_lock |= (1 << k);
        }
    }
    for(uint8_t i = 0; i < ROW; i++)
    {
        *outdata |= (1 << (3 * i));
    }
}

I have another file, called datafield.h, what consists the matrix:

static const uint8_t signal_table[][COLS] =
{        
    {22, 24, 63, 67}, // 1A
    {87, 89, 31, 35}, // 1B
    {0 ,  2, 21, 24}, // 1C
    {0 ,  3, 22, 43}, // 1D
}

In the main I need to refer to the function like this, to avoid errors and warnings:

int main(void)
{
    initLamp(0, *signal_table, &returndata);
}

The question is: I'm a bit sceptic using the main function. I'm almost sure that the usage of '&returndata' is correct, but why do I have to put the '*' operator before the 'signal_table'. I thought I would only need to give the address of the matrix like &signal_table, but with this it is not working.

signal_table is of type const uint8_t [4][4]. As an operand of unary *, signal_table is converted from const uint8_t [4][4] to const uint8_t (*)[4] (pointer to array length 4 of const uint8_t), so *signal_table is of type const uint8_t [4]. As a function call argument, *signal_table is converted from const uint8_t [4] to const uint8_t *, which is compatible with the function parameter. It would be more natural to replace *signal_table with &signal_table[0][0], which is also of type const uint8_t * and points to the same thing. — Ian Abbott
– Ian Abbott, Commented Nov 23, 2021 at 13:27
In general, using pointer arithmetic on a pointer to some element to access an element in a different row of a 2-D array (treating a 2-D array as a flat 1-D array) leads to undefined behavior. There is an exception for pointer to character types (char, signed char, unsigned char) which can legitimately be used to access any byte of the 2-D array. It is possible that uint8_t is not defined as a character type on some C implementation (it could be defined as an extended integer type), which would disallow this usage. If uint8_t is the same type as unsigned char then it is OK. — Ian Abbott
– Ian Abbott, Commented Nov 23, 2021 at 14:45

Lundin · Accepted Answer · 2021-11-23 13:03:31Z

2

Avoid writing needlessly obscure or complicated code when you could keep it simple. Your function should probably be written as:

void initLamp(uint8_t time, const uint8_t datatable[ROWS][COLS], uint64_t *outdata)

Or alternatively with VLA syntax:

void initLamp(uint8_t time, 
              size_t rows, 
              size_t cols, 
              const uint8_t datatable[rows][cols], 
              uint64_t *outdata)

In case of the latter, call it as:

initLamp(0, ROWS, COLS, signal_table, &returndata);

And the for loops should be simplified to use array syntax:

for(size_t r = 0; r < rows; r++)
{
  for(size_t c = 0; c < cols; c++)
  {
    datatable[r][c] = something;
  }
}

Another issue is the 1 << ... parts of your code, which are likely hidden bugs. 1 is an integer constant of type int. You might end up shifting data into the sign bit which is an undefined behavior bug. In case of uint64_t you should rather be using 1ULL or (uint64_t)1.

answered Nov 23, 2021 at 13:03

Lundin

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1 Comment

T.Barta Over a year ago

Thanks for the tip, especially for the hidden bug, since I was struggling with it for some minutes.

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