无法理解c中的2d数组中的寻址

Question

main()
{
    int a[3][2] = { {1,2},{3,4},{5,6}};
    for(int i=0;i<3;i++)
        for(int j=0;j<2;j++)
    {
        printf("%d", a[i][j]);
        printf("\t %d\n", &a[i][j]);
    }
    printf("\n%d", *(a+1));
    printf("\n%d", *a+1);
}

the output of *(a+1) is different from *a+1.

*(a+1) is pointing to 3 rd element whereas

*a+1 is outputting the 2nd value

Answer 1

the output of *(a+1) is different from *a+1.

Yes, due to operator precedence . *a + 1 means...

Dereference a (which returns an int ) and add 1 to it. Return the result ( 2 )

However, *(a + 1) says...

Add 1 to the pointer a and dereference it, ie, get the value at the address a + sizeof(int[2]) .

The "value* happens to be the first element of the second array. Remember; adding n to a pointer type advances the address by n elements . In this case, each element is an array of int with two elements of its own.

That should answer the next two questions as well. After reading up on operator precedence, start studying pointer arithmetic .

Answer 2

C doesn't really have multi-dimensional arrays, it has arrays of arrays. int a[3][2] declares a 3-element array whose elements are 2-dimensional arrays of ints.

Arithmetic on pointers takes this into account, it increments by the size of the objects that the pointer points to. So (a+1) evaluates to the second element of a , which is a pointer to the array {3, 4} . Indirecting through this pointer gets you its first element, which is 3.

*a on the other hand, indirects through pointer to the first element of a , which evaluates to 1. Then when you add 1 to this, you get 2. It's not actually returning the 2nd element, it just looks like it because 1+1 = 2. Try changing your initialization to:

int a[3][2] = { {1,3},{5,7},{9,11}};

and your output will be 5 and 2.

Answer 3

You should read about operator precedence. In the second case *a+1 the unary operator * binds more than the binary operator + . So, the operation is treated like adding one to the value pointed by a .

In the former case *(a+1) , you are addressing the next element pointed by a and dereferencing it to get its value.

HTH.

Answer 4

Yes,that's correct. Because a 2D array represent an array of 1D arrays.In this case in a[3][2] = { {1,2},{3,4},{5,6}} ,name of array ,that is a acts as zeroth element of array. Since zeroth element of our 2-D array is 1-D array of 2 integers{1,2}.Thus, (a+1) refers to address of first element of 2D array {3,4} and *(a+1) thus gives value at that address ie 3. While,*a gives access to first element and than it increments it by one giving 2.