Can sizeof return 0 (zero)

Question

Is it possible for the sizeof operator to ever return 0 (zero) in C or C++? If it is possible, is it correct from a standards point of view?

Answer 1

In C++ an empty class or struct has a sizeof at least 1 by definition. From the C++ standard, 9/3 "Classes": "Complete objects and member subobjects of class type shall have nonzero size."

In C an empty struct is not permitted, except by extension (or a flaw in the compiler).

This is a consequence of the grammar (which requires that there be something inside the braces) along with this sentence from 6.7.2.1/7 "Structure and union specifiers": "If the struct-declaration-list contains no named members, the behavior is undefined".

If a zero-sized structure is permitted, then it's a language extension (or a flaw in the compiler). For example, in GCC the extension is documented in "Structures with No Members", which says:

GCC permits a C structure to have no members:

 struct empty {
 };

The structure will have size zero. In C++, empty structures are part of the language. G++ treats empty structures as if they had a single member of type char.

Answer 2

sizeof never returns 0 in C and in C++. Every time you see sizeof evaluating to 0 it is a bug/glitch/extension of a specific compiler that has nothing to do with the language.

Answer 3

Every object in C must have a unique address. Worded another way, an address must hold no more than one object of a given type (in order for pointer dereferencing to work). That being said, consider an 'empty' struct:

struct emptyStruct {};

and, more specifically, an array of them:

struct emptyStruct array[10];
struct emptyStruct* ptr = &array[0];

If the objects were indeed empty (that is, if sizeof(struct emptyStruct) == 0), then ptr++ ==> (void*)ptr + sizeof(struct emptyStruct) ==> ptr, which doesn't make sense. Which object would *ptr then refer to, ptr[0] or ptr[1]?

Even if a structure has no contents, the compiler should treat it as if it is one byte in length in order to maintain the "one address, one object" principle.

The C language specification (section A7.4.8) words this requirement as

when applied to a structure or union, the result (of the sizeof operator) is the number of bytes in the object, including any padding required to make the object tile an array

Since a padding byte must be added to an "empty" object in order for it to work in an array, sizeof() must therefore return a value of at least 1 for any valid input.

Edit: Section A8.3 of the C spec calls a struct without a list of members an incomplete type, and the definition of sizeof specifically states (with emphasis added):

The operator (sizeof) may not be applied to an operand of function type, or of incomplete type, or to a bit-field.

That would imply that using sizeof on an empty struct would be equally as invalid as using it on a data type that has not been defined. If your compiler allows the use of empty structs, be aware that using sizeof on them is not allowed as per the C spec. If your compiler allows you to do this anyway, understand that this is non-standard behavior that will not work on all compilers; do not rely on this behavior.

Edit: See also this entry in Bjarne Stroustrup's FAQ.

Answer 4

Empty structs, as isbadawi mentions. Also gcc allows arrays of 0 size:

int a[0];
sizeof(a);

EDIT: After seeing the MSDN link, I tried the empty struct in VS2005 and sizeof did return 1. I'm not sure if that's a VS bug or if the spec is somehow flexible about that sort of thing

Answer 5

typedef struct {
  int : 0;
} x;

x x1;
x x2;

Under MSVC 2010 (/Za /Wall):

sizeof(x) == 4
&x1 != &x2

Under GCC (-ansi -pedantic -Wall) :

sizeof(x) == 0
&x1 != &x2

i.e. Even though under GCC it has zero size, instances of the struct have distinct addresses.

ANSI C (C89 and C99 - I haven't looked at C++) says "It shall be possible to express the address of each individual byte of an object uniquely." This seems ambiguous in the case of a zero-sized object, since it arguably has no bytes.

Edit: "A bit-field declaration with no declarator, but only a colon and a width, indicates an unnamed bit-field. As a special case of this, a bit-field with a width of 0 indicates that no further bit-field is to be packed into the unit in which the previous bit-field, if any, was placed."

Answer 6

in my view, it is better that sizeof returns 0 for a structure of size 0 (in the spirit of c). but then the programmer has to be careful when he takes the sizeof an empty struct.

but it may cause a problem. when array of such structures is defined, then

&arr[1] == &arr[2] == &arr[0]

which makes them lose their identities.

i guess this doesnt directly answer your question, whether it is possible or not. well that may be possible depending on the compiler. (as said in Michael's answer above).

Answer 7

I think it never returns 0 in c , no empty structs is allowed

Answer 8

Here's a test, where sizeof yields 0

#include <stdio.h>

void func(int i)
{
        int vla[i];
        printf ("%u\n",(unsigned)sizeof vla);
}


int main(void)
{

        func(0);
        return 0;
}

Answer 9

If you have this :

struct Foo {}; 
struct Bar { Foo v[]; }

g++ -ansi returns sizeof(Bar) == 0. As does the clang & intel compiler.

However, this does not compile with gcc. I deduce it's a C++ extension.

Answer 10

struct Empty {
} em;

struct Zero {
    Empty a[0];
} zr;

printf("em=%d\n", sizeof(em));
printf("zr=%d\n", sizeof(zr));

Result:

em=1
zr=0