I have two arrays
int x[]={1,2,3,4};
int y[]={5,6,7};
I like to create two dimensional array. assign x and y. Can I do something like this
int x[]={1,2,3,4};
int y[]={5,6,7};
int *k=x;
int *l=y;
int **m={k,l};
printf("%d\n",m[0][0]);
but at printf when 1 is get accessed throws segFault
What I am doing wrong
int **m={k,l};
is wrong. This is assigning the value of k (int*) to the variable m (int**) and leaving the excess element l in the initializer unused.
You can use an array
int *m[]={k,l};
or a compound literal
int **m=(int*[]){k,l};
instead.
Arrays are not pointers. Pointers are not arrays. And therefore pointer-to-pointers are not arrays of arrays either.
Arrays may in some circumstances "decay" into pointers to their first element, but that doesn't make them pointers.
Correct code in your case is:
int x[]={1,2,3,4};
int y[]={5,6,7};
int xy[4][3];
memcpy(xy, x, 4*sizeof x);
memcpy(xy, y, 3*sizeof y);
Or alternatively allocate the 2D array dynamically:
int (*xy)[3] = malloc( sizeof(int[4][3]) );
If it makes sense, you could create a pointer-based look-up table using pointers:
int* table [2] = {x, y};
But this will likely have slower access time than a 2D array.
int **m={k,l};. Your code should work if you change that line toint *m[2]={k,l};, makingman array and not a pointer; this could (and should, for robustness) be changed toint *m[]={k,l};. The latter deduces the (still fixed at compile time) array size from the number of items between the braces, so you don't have to change a number if in the future you add another array element. But it is still a true array with two elements and not a pointer likeint **mis.