C - 'char **' differs in levels of indirection from 'char (*)[6]'

Question

Can someone please explain to me what\'s wrong with the following, and more importantly why?

int main( int argc, char *argv[] )
{
    char array[] = \"array\         


        

Answer 1

This would be the way to do what you seem to be trying to do:

int main( int argc, char *argv[] )
{
  char array[] = "array";
  char (*test)[6];
  test = &array;

  (*test)[0] = 'Z';

  printf( "%s\n", array );

  return 0;
}

test is a pointer to an array, and an array is different from a pointer, even though C makes it easy to use one like the other in my cases.

If you wanted to avoid having to specify a specific sized array, you could use a different approach:

int main( int argc, char *argv[] )
{
  char array[] = "array";
  char *test;
  test = array;  // same as test = &array[0];

  test[0] = 'Z';

  printf( "%s\n", array );

  return 0;
}

Answer 2

char **test; is a pointer to a pointer, but if you're going to take the address of an entire array then it needs to be a pointer to entire array i.e char (*test)[6];

Answer 3

test = &array

is wrong because test is of type char** and &array is a char(*)[6] and is a different type from char**

An array isn't the same type as char* although C will implicitly convert between an array type and a char* in some contexts, but this isn't one of them. Basically the expectation that char* is the same as the type of an array (e.g: char[6]) is wrong and therefore the expectation that taking the address of an array will result in a char** is also wrong.