[Solved] what is the output? Please explain, considering i am a novice in c [closed]

Question

Time for a crash course on arrays in C.

First of all, let’s fix the initializer for the array:

int a[3][4] = {
                { 1,  2,  3,  4},
                { 5,  6,  7,  8},
                { 9, 10, 11, 12}
              };

This defines a 3-element array of 4-element arrays of int. The type of the expression a is “3-element array of 4-element arrays of int“.

Now for the headache-inducing part. Except when it’s the operand of the sizeof or unary & operators, or if it’s a string literal being used to initialize another array in a declaration, an expression of array type will have its type implicitly converted (“decay”) to a pointer type.

If the expression a appears by itself in the code (such as in a statement like printf("%p", a);, its type is converted from “3-element array of 4-element array of int” to “pointer to 4-element array of int“, or int (*)[4]. Similarly, if the expression a[i] appears in the code, its type is converted from “4-element array of int” (int [4]) to “pointer to int” (int *). If a or a[i] are operands of either sizeof or &, however, the conversion doesn’t happen.

In a similar vein, array subscripting is done through pointer arithmetic: the expression a[i] is interpreted as though it were written *(a+i). You offset i elements from the base of the array and dereference the result. Thus, a[0] is the same as *(a + 0), which is the same as *a. a[i][j] is the same as writing *(*(a + i) + j).

Here’s a table summarizing all of the above:

Expression   Type          Decays To    Resulting Value
----------   ----          ---------    -----
         a   int [3][4]    int (*)[4]   Address of the first element of the array
        &a   int (*)[3][4] n/a          Same as above, but type is different
        *a   int [4]       int *        Same as above, but type is different
      a[0]   int [4]       int *        Same as above
    *(a+0)   int [4]       int *        Same as above
      a[i]   int [4]       int *        Address of the first element of the i'th subarray
    *(a+i)   int [4]       int *        Same as above
     &a[i]   int (*)[4]    n/a          Same as above, but type is different
     *a[i]   int           n/a          Value of the 0'th element of the i'th subarray
   a[i][j]   int                        Value of the j'th element of the i'th subarray
 *(a[i]+j)   int                        Same as above
*(*(a+i)+j)  int                        Same as above

Hopefully, that should give you everything you need to figure out what the output should be. However, the printf statement should be written as

printf("%p %d %d\n", (void *) a[0]+1, *(a[0]+1), *(*(a+0)+1));

Accepted Answer