Clarify array of structs pointers in C

Question

I have a generic struct declared and an array of these structs as given below:

struct A
{
    int x,y,z;
    char a,b,c;
};

struct A *str_arr[5];

From my understanding str_arr is a pointer to a block of memory which stores pointers to the 5 structs in sequential order and therefore these pointers can be accessed via pointer arithmetic or array indexing as:

struct A *str_a = str_arr[1];                    // points to 2nd struct?
struct A *str_b = str_arr + 2*sizeof(struct A*); // points to 3rd struct?

However, these 5 structs might not be in sequential memory?

printf("%p\n", str_arr);              // prints memory location of start of str_arr pointers?
printf("%p\n", str_arr[1])            // prints memory location of 2nd struct?
printf("%d\n" str_arr == &str_arr[0]) // prints 1?

I would just like clarification that my understanding is correct with all of the points I have raised.

str_arr is array of pointer to struct A (five pointer to struct A). — BLUEPIXY
– BLUEPIXY, Commented May 28, 2015 at 5:45
Just like the integers in int a[42] do not have to be in sequential order. They are whatever you put in there. — juanchopanza
– juanchopanza, Commented May 28, 2015 at 5:57

Mohit Jain · Accepted Answer · 2015-05-28 05:56:04Z

4

All is correct except one:

struct A **str_b = str_arr + 2 /* *sizeof(struct A*) */;
/*       ^^                    ^^^^^^^^^^^^^^^^^^^^^^ */
/* Not need to multiply with size. Dereference with * if your type is struct A * */

or

struct A *str_b = *(str_arr + 2);

You give the offset in terms of number of elements and not the size in bytes.

str_arr + 2*sizeof(struct A*) is equivalent to &str_arr[2*sizeof(struct A*)]

             +0  +1  +2  +3  +4
           +---+---+---+---+---+
           | A | B | C | D | E |
           +---+---+---+---+---+
str_arr    ^^^^^^^^^^^^^^^^^^^^^
&str_arr[0]^^^^   
str_arr[1] = B

str_arr is the address of array start
str_arr[1] is the contents at offset +1 i.e. B which is an address pointing to object of type struct A.
str_arr == &str_arr[0] are same address with different type

As suggested by @Gopi, last point can be proven by printing the following:

sizeof str_arr v/s &str_arr[0]
Following address, &str_arr + 1 v/s str_arr + 1

edited May 28, 2015 at 5:56

answered May 28, 2015 at 5:45

Mohit Jain

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3 Comments

Mohit Jain Over a year ago

Yes, it does, only with the different type. That means address generated from &str_arr and str_arr + 0 are same but addresses generated from expressions &str_arr + 1 and str_arr + 1 are different

Gopi Over a year ago

Yeah I know I meant that also can be added to the answer

Mohit Jain Over a year ago

@Gopi Thanks for the input. I thought it is implied by the figure. Length of ^^^is different for str_arr and &str_arr[0]. But still feel free to edit the answer if you can add more helpful or easier to read info for OP. I will try to add your suggestion.

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