C: How do I correctly access an Array element of a Struct passed by reference?

Question

I have this code:

typedef struct _structVar{
   int myArray[10];
} structVar;

structVar MyStruct;

Then I'm passing the struct by reference to a function:

myFunct(&MyStruct);

How I access array elements inside MyFunct?

void myFunct(structVar *iStruct){

   for(char i=0; i<10; i++)
      (*iStruct)->myArray[i] = i; //Fix Here
}

Thanks for the help.

`*iStruct` is the structure object itself, not a pointer to the structure. The "arrow" operator `->` only operates on pointers (it implies dereference). — Some programmer dude, Jul 22 '19 at 09:58
And please remember that C doesn't have pass-by-reference at all, only pass-by-value. You pass the pointer by value, which in a way can be seen as *emulating* pass by reference, but it's still not proper reference passing. — Some programmer dude, Jul 22 '19 at 10:00
@Someprogrammerdude: Emulating pass by reference is pass by reference. C++ adopted a built-in pass-by-reference, but that does not change the fact that if you pass by value a thing X that refers to a thing Y by pointing to it, you have passed a reference to Y. — Eric Postpischil, Jul 22 '19 at 10:51

score 5 · Accepted Answer · answered Jul 22 '19 at 10:02

5

Either write

iStruct->myArray[i] = i;

or

(*iStruct).myArray[i] = i;

answered Jul 22 '19 at 10:02

Vlad from Moscow

You should mention that the first way is the better (more usual) way. – Jabberwocky Jul 22 '19 at 10:08
@Jabberwocky I consider them as equivalent.:) – Vlad from Moscow Jul 22 '19 at 10:10
Of course they are equivalent, but (almost) nobody uses the second ways. – Jabberwocky Jul 22 '19 at 10:12

1 Answers1