In unqlite c library I found following code:
pObj = jx9VmReserveMemObj(&(*pVm),&nIdx);
where pVm
is:
typedef struct jx9_vm jx9_vm;
jx9_vm *pVm
and function called is declared as:
jx9_value * jx9VmReserveMemObj(jx9_vm *, sxu32 *);
What for construct &(*pVm)
is used in call instead of just pVm
? Is &(*pVm)
equivalent to pVm
?
Quoting C11
, chapter §6.5.3.2, Address and indirection operators
[...] If the operand is the result of a unary
*
operator, neither that operator nor the&
operator is evaluated and the result is as if both were omitted, except that the constraints on the operators still apply and the result is not an lvalue. [...]
So, yes, they are equivalent.
This construct can be used, however, to check the type of the argument against a pointer type. From the property of unary *
operator,
The operand of the unary
*
operator shall have pointer type.
So, the construct &(*pVm)
pvm
is a pointer or array name.pvm
is a non-pointer type variable.See the other answer by Alter Mann for code-wise example.
One more difference (in general) is, pVm
can be assigned (can be used as LHS of the assignment operator), but &(*pVm)
cannot.
Is
&(*pVm)
equivalent topVm
?
Yes. *1
Same for *(&pVm)
.
(*1)
As the *
-operator (de-referencing) only is applicable to pointers, the former construct only works on a pointer (or an array, which would decay to a pointer to its 1st element). The latter can be applied to any type of variable.:
Yes, they are the same, but notice that it fails when the object is not an array or a pointer:
#include <stdio.h>
struct t {
int value;
};
typedef struct t t;
int func(t *ptr)
{
return ptr->value;
}
int main(void)
{
t o = {.value = 0};
t v[2] = {{.value = 1}, {.value = 2}};
printf("%d\n", func(&(*o))); /* error: invalid type argument of unary ‘*’ */
printf("%d\n", func(&(*v))); /* WORKS */
return 0;
}
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