pointer to array of structs

What is the proper syntax for initializing a pointer to an array of structures?

I have the following definitions:

Code:

typedef struct
{
   int Targets;
   Target_t targets[TARGETS_MAX];
} TargetList_t;

typedef struct  
{
  int id;                           
  float distX;                        
  float distY;                       

}Target_t;

TargetList_t foo[SENS_MAX];
TargetList_t* ptrfoo = &foo;
TargetList_t TL;
TargetList_t* ptrTL = &TL;

If I try to initialize the pointer ptrTL it works fine

Code:

ptrTL->targets[0].distX = 4;

However, in initializing the pointer to foo

Code:

ptrfoo[0]->targets[0].distX = 2;

I get an error saying:

error C2232: '->targets' : left operand has 'struct' type, use '.'

what gives? How can I initialize a pointer to an array of structures?

For one, this is wrong:

Code:

TargetList_t foo[SENS_MAX];
TargetList_t* ptrfoo = &foo;

Since foo is an array, "foo" by itself is a pointer. (aka, foo == &foo[0]).

Todd

> TargetList_t* ptrfoo = &foo;
You don't need the & in this assignment.

> ptrfoo[0]->targets[0].distX = 2;
A pointer to an array, and an array share the same notation, so in this case it would be
ptrfoo[0].targets[0].distX = 2;

Being the first element of the array, you could have done this as well
ptrfoo->targets[0].distX = 2;

Originally Posted by Todd Burch

For one, this is wrong:
Since foo is an array, "foo" by itself is a pointer. (aka, foo == &foo[0]).

Todd

Ahh, I did not know that. Thanks for the tip.

You should have gotten a warning on that.
Conversion from TargetList_t** to TargetList_t*.
Do you have warnings enabled?