I am a teaching assistant of a introductory programming course, and some students made this type of error:
char name[20];
scanf("%s",&name);
which is not surprising as they are learning... What is surprising is that, besides gcc warning, the code works (at least this part). I have been trying to understand and I wrote the following code:
void foo(int *v1, int *v2) {
if (v1 == v2)
printf("Both pointers are the same\n");
else
printf("They are not the same\n");
}
int main() {
int test[50];
foo(&test, test);
if (&test == test)
printf("Both pointers are the same\n");
else
printf("They are not the same\n");
}
Compiling and executing:
$ gcc test.c -g
test.c: In function ‘main’:
test.c:12: warning: passing argument 1 of ‘foo’ from incompatible pointer type
test.c:13: warning: comparison of distinct pointer types lacks a cast
$ ./a.out
Both pointers are the same
Both pointers are the same
Can anyone explain why they are not different?
I suspect it is because I cannot get the address of an array (as I cannot have & &x
), but in this case the code should not compile.
Edit: I know that an array by itself is the same as the address of the first element, but this is not related to this problem, I think. For example:
int main() {
int a[50];
int * p = a;
printf("%d %d %d\n", p == a, p == &a[0], &p[0] == a);
printf("%d %d %d\n", p == &a, &p == a, &p == &a);
}
prints:
$ ./a.out
1 1 1
1 0 0
I don't understand why the second line begins with 1
.
Actually, they are different, they don't have the same type at least.
But in C, the address of the array is the same as the address of the first element in the array that's why "they are not different", basically, they point to the same thing.
The name of an array, in most circumstances, evaluates to the address of its initial element. The two exceptions are when it is the operand of
sizeof
or the unary&
.The unary
&
gives the address of its argument. The address of an array is the same as the address of its initial element, so(void*)&array == (void*)array
will always be true.array
, when converted to a pointer to its initial element, has the typeT *
. The type of&array
isT (*)[n]
, wheren
is the number of elements in the array. Thus,I believe this is a gcc optimization. Think about it.
&test
points to the address oftest
test
points to the first element oftest
or&test[0]
[0]
is the same(for the most part) as*
So according to this
&test
could be different thantest
but gcc optimizes this away because there is no purpose of having an extra level of indirection at that point.In your example, the array
test
is a block of 50ints
. So it looks like this:When you apply the unary
&
operator to an array, you get the address of the array. Just like when you apply it to anything else, really. So&test
is a pointer that points to that block of 50ints
:A pointer that points to an array of 50 ints has type
int (*)[50]
- that's the type of&test
.When you just use the name
test
in any place where it's not the operand of either thesizeof
or unary-&
operators, it is evaluated to a pointer to its first element. So thetest
that you pass tofoo()
evaluates to a pointer to thetest[0]
element:You can see that these both are pointing to the same address - although
&test
is pointing to the whole array, andtest
is pointing to the first element of the array (which only shows up in the different types that those values have).If You define an array like
name
is implicitly convertible tochar*
, but&name
is of the typechar (*)[20]
(a pointer to an array of 20 characters). The addresses are the same.Check the address of
(&name + 1)
. It differs form&name
by thesizeof(char [20])
.