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;
}