getting stuck with C pointer Structure

Question

I am working on Embedded C . I am stuck with pointer structure ....

Structures are given below ..

/*structure 1*/
ZPS_tsAplApsmeBindingTableType *psAplApsmeAibBindingTable;

/*structure 2*/
typedef struct
{
    ZPS_tsAplApsmeBindingTableCache* psAplApsmeBindingTableCache;
    ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable;
}ZPS_tsAplApsmeBindingTableType;

/*structure3*/
typedef struct
{
   uint64  u64SourceAddress;
   ZPS_tsAplApsmeBindingTableEntry* pvAplApsmeBindingTableEntryForSpSrcAddr;
   uint32 u32SizeOfBindingTable;
}ZPS_tsAplApsmeBindingTable;

/*structure 4*/
typedef struct
{
   ZPS_tuAddress  uDstAddress;
   uint16  u16ClusterId;
   uint8   u8DstAddrMode;
   uint8   u8SourceEndpoint;
   uint8   u8DestinationEndPoint;
} ZPS_tsAplApsmeBindingTableEntry;

I have declared ZPS_tsAplApsmeBindingTableType *p; but i want to access ZPS_tsAplApsmeBindingTableEntry structure values... How could i do that??

Can anyone tell me the difference between

ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable

and

ZPS_tsAplApsmeBindingTable *psAplApsmeBindingTable;

Thanks ....

I'm stuck on your naming conventions! Wow. – Duck Feb 20 '12 at 18:55 — Duck, Feb 20 '12 at 18:55

score 4 · Accepted Answer · answered Feb 20 '12 at 18:57

4

p->psAplApsmeBindingTable->pvAplApsmeBindingTableEntryForSpSrcAddr->someField
No difference.

PS. That code is really, really ugly.

answered Feb 20 '12 at 18:57

James M

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and I think ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable is a array of structure ...if i use psAplApsmeBindingTable[0].someval or psAplApsmeBindingTable->someptr its showing a illegal address alignment – Shantanu Banerjee Feb 21 '12 at 03:37

score 2 · Answer 2 · answered Feb 20 '12 at 18:59

2

ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable;

and

ZPS_tsAplApsmeBindingTable *psAplApsmeBindingTable;

and

ZPS_tsAplApsmeBindingTable * psAplApsmeBindingTable;

Are the same. The location of the * does not change anything.

To access values of a struct pointed to by a pointer (like your pointer p), you can use an arrow ->

p->psAplApsmeBindingTable->pvAplApsmeBindingTableEntryForSpSrcAddr->u16ClusterId

answered Feb 20 '12 at 18:59

Sam DeHaan

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but sir it's show me error .. ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable its a array of structure pointed by psAplApsmeBindingTable.. if i use p->psAplApsmeBindingTable[0].u64SourceAddress again its show error.. please suggest a way to solve it.. – Shantanu Banerjee Feb 21 '12 at 03:42
have you tried `p->psAplApsmeBindingTable[0]->u64SourceAddress ` ? – Sam DeHaan Feb 21 '12 at 13:50

score 1 · Answer 3 · answered Feb 20 '12 at 18:55

I have declared ZPS_tsAplApsmeBindingTableType *p; but i want to access ZPS_tsAplApsmeBindingTableEntry structure values... How could i do that??

Well you can't. In your code a ZPS_tsAplApsmeBindingTableType does not contain any members of the type ZPS_tsAplApsmeBindingTableEntry or ZPS_tsAplApsmeBindingTableEntry*.

Can anyone tell me the difference between ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable and ZPS_tsAplApsmeBindingTable *psAplApsmeBindingTable;

There is no difference; they are the same thing... literally the same text copied twice. I don't really understand your question. If you could elaborate a bit I may be able to help further.

suppose when a array of structure is pointed by a structure pointer ZPS_tsAplApsmeBindingTable* psAplApsmeBindingTable its a structure pointer .... i want to access those values in that location ... — Shantanu Banerjee, Feb 21 '12 at 03:32

score 0 · Answer 4 · answered Feb 20 '12 at 20:13

You'd get to the ZPS_tsAplApsmeBindingTableEntry members as follows:

p->psAplApsmeBindingTable->pvAplApsmeBindingTableEntryForSpSrcAddr->uDstAddress
p->psAplApsmeBindingTable->pvAplApsmeBindingTableEntryForSpSrcAddr->u16ClusterId
p->psAplApsmeBindingTable->pvAplApsmeBindingTableEntryForSpSrcAddr->u8DstAddrMode

etc. You would use the -> for all selections because p, psAplApsmeBindingTable, and pvAplApsmeBindingTableEntryForSpSrcAddr are all pointers to struct types. If any of them had not been pointer types, you would have used the . operator to do the component selection for that type. For example:

struct a {int x; int y};
struct b {struct a *p; struct a v};
struct b foo, *bar = &foo;
...
foo.p->x = ...;  // foo is not a pointer type, p is a pointer type
bar->v.y = ...;  // bar is a pointer type, v is not a pointer type

The expression x->y is shorthand for (*x).y; IOW, you dereference x and then select y.

There is no difference between the declarations

T *p;

and

T* p;

Both are interpreted as T (*p); - the * is always part of the declarator, not the type specifier. If you wrote

T* a, b;

only a would be declared as a pointer to T; b would be declared as a plain T.

getting stuck with C pointer Structure

4 Answers4