cannot convert argument of incomplete type 'void *' to 'const bitData'

Question

I am trying to create a bit array in c++ and it works when I create a single variable. But I can't figure out how to fix this error when I am trying to create an array out of it.

typedef struct bitData
{
      unsigned bit:1;
} bitData;

int main() {
  bitData* buff;
  *buff = malloc(3 * sizeof(struct bitData));
  buff[0].bit = 1;
  buff[1].bit = 0;
  buff[2].bit = 1;

  for (int i = 0; i < 3; ++i)
  {
    printf("buff[%d] = %d\n",i , buff[i].bit);
  }
}

Here is the error :

$ g++ main.cpp 
main.cpp:23:9: error: no viable overloaded '='
  *buff = malloc(3* sizeof(struct bitData));
  ~~~~~ ^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.cpp:12:16: note: candidate function (the implicit copy assignment operator) not viable: cannot convert argument of incomplete type 'void *' to 'const bitData'
typedef struct bitData
          ^
1 error generated.

@PierreEmmanuelLallemant .. I tried that but it give `error: assigning to 'bitData *' from incompatible type 'void *'` — molecule, Oct 21 '16 at 08:12
Note that this is probably the most inefficient way to have an array of bits. Instead of `malloc`, you should look for `std::vector`, and for array of bits `std::vector` specifically. What you are doing now is writing C style code in C++, which is far from ideal. — user694733, Oct 21 '16 at 08:21
Note that each one of your `bitData` is the size of an `unsigned int`, most likely 32 bits. — molbdnilo, Oct 21 '16 at 08:24
@molecule Nothing can be smaller than a `char`. There are numerous articles and tutorials online about manipulating bits. — molbdnilo, Oct 21 '16 at 08:37

nvoigt · Answer 1 · 2016-10-21T08:17:36.500

3

There is a typo in your code. The * in front of your pointer variable needs to go, you don't want to dereference it here, you want to assign to it.

In C:

int main() {
  bitData* buff;
  buff = malloc(3 * sizeof(struct bitData));

In a horrible C/C++ hybrid:

int main() {
  bitData* buff;
  buff = (bitData*)malloc(3 * sizeof(struct bitData));

In C++:

int main() {
  bitData* buff = new bitData[3];

edited Oct 21 '16 at 08:17

answered Oct 21 '16 at 08:12

nvoigt

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That's still not valid C++ without a cast – krzaq Oct 21 '16 at 08:13
yeah .. although you are right but that is still not valid – molecule Oct 21 '16 at 08:15
@krzaq That's why I asked him whether this is supposed to be C or C++... [In C you don't](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). – nvoigt Oct 21 '16 at 08:15
@nvoigt the question text says C++, tags say C++ and C. So it's ambiguous but rather C++. – krzaq Oct 21 '16 at 08:17
1

@krzaq Well, the code itself is plain C. But I expanded my answer to contain other options. – nvoigt Oct 21 '16 at 08:18

score 2 · Accepted Answer · answered Oct 21 '16 at 08:16

*buff is an lvalue of type bitData. You can't assign a pointer to it.

In C, you can just say this:

buff = malloc(3* sizeof(struct bitData));

In C++ you cannot directly assign a void* to T*, you need to cast it:

buff = static_cast<bitData*>(malloc(3* sizeof(struct bitData)));

Although if you're using C++ you should use new rather than malloc, or, even better, containers/smart pointers to avoid manually managing memory.

If you want this part of code compatible with both C and C++, use C-style cast (be sure you really need to do this, though):

buff = (bitData*)malloc(3* sizeof(struct bitData));

Then follow @user694733's advice. `vector` mayb be exactly what you want. What you're doing here is allocating 3 bytes, for 3 structs, even if you only use 1 bit of each. — krzaq, Oct 21 '16 at 08:30

score 0 · Answer 3 · answered Oct 21 '16 at 08:14

Just change the code as below and try:

typedef struct bitData
{
      unsigned bit:1;
} bitData;

int main() {
  bitData* buff;
  buff = (bitData*)malloc(3 * sizeof(struct bitData));
  buff[0].bit = 1;
  buff[1].bit = 0;
  buff[2].bit = 1;

  for (int i = 0; i < 3; ++i)
  {
    printf("buff[%d] = %d\n",i , buff[i].bit);
  }
}

Remove the * before buff and typecast it while allocating the memory. The above code will get compiled without any error or warning.

cannot convert argument of incomplete type 'void *' to 'const bitData'

3 Answers3