Question

I am getting confused as to whether it is valid (in C) to pass a pointer to an array that has been initiated as follows (e.g. at compile time on the stack):

int a[] = {1, 2, 3};

my_func(&a);


void my_func(int **a)
{
   int *tmp = *a; /* this statement fixes the problem */
   printf("%d %d %d", (*a)[0], (*a)[1], (*a)[2]); /*doesn't work */ 
   printf("%d %d %d", tmp[0], tmp[1], tmp[2]); /*does work */ 
}

when I step through this with gdb I can’t ‘see’ any of the values (*a)[0], etc. from ‘inside’ my_func. e.g.

(gdb) p (*a)[0]
Cannot access memory at address 0x0

I’m thinking that possibly I have a fundamental misunderstanding with regard to what you can and can’t do with arrays that are on the stack rather than the heap?

I hope thats not the case as it is very convenient for my unit tests to declare arrays on the stack as in the example, but I need to test functions that are expecting pointers to pointers to int.

Note I do get a compiler warning as follows:

 test_utility.c:499:5: warning: passing argument 1 of ‘int_array_unique’ from incompatible pointer type [enabled by default]
 ../src/glamdring2.h:152:5: note: expected ‘int **’ but argument is of type ‘int (*)[12]’

but I thought it would be ok to mix int *a[] with **a? Perhaps not? Are they not equivalent?

Answer 1

a[] is an array, not a pointer (“not an lvalue”); in your function call

func( &a);

&a decays to a pointer to int; &a is not a pointer to pointer to int. Why? there is no pointer to point to.

The function prototype

void func( int **p);

expects a pointer to pointer to int, that does not fit the function being called with a pointer to int as an argument, like you did.

UPDATE: I don’t know what the OP’s intentions were, so this is just a guess…

void my_func(int *a);

int a[] = {1, 2, 3};

my_func(a); /* note: this is equivalent to my_func( &a ); */


void my_func(int *a)
{
   printf("%d %d %d\n", a[0], a[1], a[2] ); 
}