我如何将二维数组分配给其他临时二维数组…？ 在C编程中

Question

Hi I am trying to store the contents of two dimensional array into a temporary array.... How is it possible... I don't want looping over here, as it would add an extra overhead.. Any pointer notation would be good.

struct bucket
{
   int nStrings;       
   char strings[MAXSTRINGS][MAXWORDLENGTH];
};

void func()
{
   char **tArray;
   int tLenArray = 0;
   for(i=0; i<TOTBUCKETS-1; i++)
   {
      if(buck[i].nStrings != 0)
      {
         tArray = buck[i].strings;
         tLenArray = buck[i].nStrings;
      }
   }
}

The error here i am getting is:-
[others@centos htdocs]$ gcc lexorder.c
lexorder.c: In function âlexSortingâ:
lexorder.c:40: warning: assignment from incompatible pointer type

Please let me know if this needs some more explanaition...

Answer 1

You can use memcpy or wrap it in a struct and use assignment on the structs, but any way you do it, you'll effectively be looping over all the elements. That's fundamental.

Answer 2

The compiler issue is that the notation array[...][...] is not the same as char * *. The char ** notation is pointer to a pointer.

For example;

char ** pA;
pA = malloc(10);

pA[0] is a pointer

in the case of array[...][...]

array[0] is not a pointer. It is a char.

If you want to do what you are trying then you you need declare strings as char ** and initialize your structure like this.

struct bucket
{
    int nStrings;       
    char** strings;
};

struct bucket buck;

buck.strings = (char **)malloc(MAXSTRINGS);

for(i = 0; i < MAXSTRINGS; i++)
    buck.strings[i] = (char*)malloc(MAXWORDLENGTH)

Answer 3

You mean cloning the array right?
Unfortunately, I don't think there is a way to clone an array in C without looping through all items in the array and making a copy of them somewhere else.
The main problem is that since they are pointers, any thing you do to simply copy the contents would just copy the addresses and you would have a reference, not a clone.