Pointer to an array of structure in C

Question

I have a requirement to use pointer to a array of structures as defined below:

typedef void (*fptr)(uint8 arg);
typedef struct sTest
{
  uint8 u8Mem1;
  uint8 u8Mem2;
  fptr fptr1;
}tTest;

Now I have define an array of this structre as follow:

tTest sTestStructure[20];

So now I want define a pointer and assign the array of stuct{sTestStructure[20]} to this pointer.Then access the member of this structure using the pointer define. So please let me know how can I define this pointer and also how to access the members using pointer.

you can use pointet *ptrsturct that is tTest *ptstruct = sTestStructure; — Seek Addo
– Seek Addo, Commented Dec 6, 2016 at 14:34
Your two choices are to use a pointer to first element (just write sTestStructure) or pointer to array &sTestStructure which has type tTest (*)[20]. The first choice is far more common in C. — Dietrich Epp
– Dietrich Epp, Commented Dec 6, 2016 at 14:37
tTest** ptr = &sTestStructure ? (litterally a pointer to an array of your structure) — Fefux
– Fefux, Commented Dec 6, 2016 at 14:39
@Fefux: wrong type - the expression &sTestStructure has type tTest (*)[20], not tTest **. — John Bode
– John Bode, Commented Dec 6, 2016 at 16:27

John Bode · Accepted Answer · 2016-12-06 16:27:16Z

Remember that unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" (T [N]) will be converted ("decay") to an expression of type "pointer to T" (T *), and the value of the expression will be the address of the first element of the array.

This naturally leads us to the following:

tTest *ptr = sTestStructure;  // note no & operator!

The expression sTestStructure is implicitly converted from type "20-element array of tTest" to "pointer to tTest", and the value of the expression is the same as &sTestStructure[0]. To access any element of sTestStructure, you could simply index into ptr as you would sTestStructure:

ptr[i].u8Mem1 = some_value;                // sTestStructure[i].u8Mem1 = some_value
printf( "%" PRIu8 "\n", ptr[j].u8Mem2 );

Remember that the subscript operation a[i] is defined as *(a + i); given the address a, offset i elements from that address and defereference the result. Thus, the [] operator implicitly dereferences ptr, which is why we use the . operator to access each struct member.

Alternately, you could also access struct members using the -> operator, and advance the pointer as necessary:

tTest *ptr = sTestStructure;
while( ptr->u8Mem1 != some_value )  // no subscript operation here
  ptr++;

The expression ptr->u8Mem1 is equivalent to (*ptr).u8Mem1¹, which is equivalent to ptr[0].u8Mem1.

So what happens if we decide to use &sTestStructure instead? Since sTestStructure is the operand of the unary & operator, the conversion rule above doesn't apply; instead of getting a pointer to a pointer to tTest, we get a pointer to a 20-element array of tTest, or:

tTest (*arrPtr)[20] = &sTestStructure;

This presents a bit more of a challenge, since we have to dereference arrPtr before we can index into it:

(*arrPtr)[i].u8Mem1 = some_value;
printf( "%" PRIu8 "\n", (*arrPtr)[j].u8Mem2 );

Since a[i] is defined as *(a + i), the expression *arrPtr can also be written as arrPtr[0] (*arrPtr == *(arrPtr + 0) == arrPtr[0]). So those lines could also be written as

arrPtr[0][i].u8Mem1 = some_value;
printf( "%" PRIu8 "\n", arrPtr[0][j].u8Mem2 );

As should be evident from those last couple of lines, you normally don't see this form of an array pointer unless you're dealing with multi-dimensional arrays. Remember our conversion rule, where an expression of type T [N] is converted to an expression of type T *? Well, replace T with an array type like Q[M], and we get the conversion from "N-element array of M-element array of Q" (Q [N][M]) to "pointer to M-element array of Q" (Q (*)[M]):

tTest sTestStructure[10][20];
tTest (*arrPtr)[20] = sTestStructure;

^{Postfix . has higher precedence than unary *, so *ptr.u8Mem1 would be parsed as *(ptr.u8Mem1), which is not what you want here.}

Juan L · Accepted Answer · 2016-12-06 14:39:11Z

0

They already told you how to define the pointer, now if you want to access a property would be *pointer->property.....

answered Dec 6, 2016 at 14:39

Juan L

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