If I have these two structs:
struct
{
int x;
} A;
struct
{
int x;
} B;
then making A = B; results in a compilation error because the two anonymous structs are not compatible.
However if I do:
typedef struct
{
int x;
} S;
S A;
S B;
A = B; is a legal assignment because they are compatible.
But why? With typedef I understand that the compiler makes this when meet S A and S B:
struct { int x; } A;
struct { int x; } B;
so A and B should not be compatible...
Each anonymous struct declaration is a distinct type; this is why you get a type mismatch when trying to assign one to the other.
A typedef, however, declares an alias (i.e. a new name for something that already exists) for a type (it does not create a new type).
A typedef is also not a simple text replacement, like a preprocessor macro. Your statement
I understand that the compiler make this when meet S A and S B:
struct { int x; } A;
struct { int x; } B;
is where your understanding is wrong.
When you use the type alias S, as in
S A;
S B;
the types of both objects A and B are the same by definition and assigning one to the other is possible.
This is because C treats every untagged struct as a new kind of struct, regardless of the memory layout. However, typedef struct { } name; cannot be used if you want to use the struct in a linked list. You'll need to stick with defining a structure tag in this case, and typedef the tagged struct instead.
struct DistanceInMeter /* Anonymous 1 */
{
int x; /* distance */
};
struct VolumeInCC /* Anonymous 2 */
{
int x; /* volume */
};
struct DistanceInMeter A;
struct VolumeInCC B;
...
A = B; /* Something is wrong here */
Equating different type doesn't always make sense and thus is not allowed.
typedef struct DistanceInMeter /* Anonymous 1 */
{
int x; /* distance */
} Dist_t;
Dist_t C, D;
...
C = D; /* Alright, makes sense */
Related
Suppose I have a struct that is defined as the following:
struct entity {
int x;
int y;
};
And a struct that uses it as a member:
struct player {
struct entity position;
char* name;
};
If I write the following code then I get an error:
struct player p;
p.x = 0; //error: 'struct player' has no member named 'x'
What I have been doing so far is writing a function that takes a player struct and returns the value by doing return player.position.x.
Is there a compiler flag, or other method, that allows me to "flatten" (I'm not sure if that's the correct phrase) the struct and allows me to access the x variable like I have shown above? I realize that this might be ambiguous if there is also an integer named x inside player as well as in entity.
Please note I will be using the entity struct in multiple structs and so I cannot use a anonymous struct inside player.
Put succinctly, the answer is "No". This is especially true if you've looked at questions such as What are anonymous structs and unions useful for in C11 and found them not to be the solution.
You can look at C11 §6.7.2.1 Structure and union specifiers for more information about structure and union types in general (and ¶13 specifically for more information about anonymous members, and ¶19 for an example). I agree that they are not what you're after; they involve a newly defined type with no tag and no 'declarator list'.
Using a macro, we can make a type generator:
#define struct_entity(...) \
struct __VA_ARGS__ { \
int a; \
int b; \
}
Then we can instantiate that type as either a tagged or anonymous structure, at will:
struct_entity(entity);
struct player {
struct_entity();
const char *name;
};
int main() {
struct player player;
player.a = 1;
player.b = 2;
player.name = "bar";
}
This code is closer in intent to what you want, and doesn't have the UB problem of the approach of declaring just the structure members in the macro. Specifically, there is a structure member inside of struct player, instead of individual members. This is important, because padding reduction and reordering of members may be performed by the compiler - especially on embedded targets. E.g. composite_1 and composite_2 below do not necessarily have the same layout!:
#include <assert.h>
#include <stddef.h>
typedef struct sub_1 {
int a;
void *b;
char c;
} sub_1;
typedef struct sub_2 {
void *d;
char e;
} sub_2;
typedef struct composite_1 {
int a;
void *b;
char c;
void *d;
char e;
} composite_1;
typedef struct composite_2 {
struct sub_1 one;
struct sub_2 two;
} composite_2;
// Some of the asserts below may fail on some architectures.
// The compile-time asserts are necessary to ensure that the two layouts are
// compatible.
static_assert(sizeof(composite_1) == sizeof(composite_2), "UB");
static_assert(offsetof(composite_1, a) == offsetof(composite_2, one.a), "UB");
static_assert(offsetof(composite_1, b) == offsetof(composite_2, one.b), "UB");
static_assert(offsetof(composite_1, c) == offsetof(composite_2, one.c), "UB");
static_assert(offsetof(composite_1, d) == offsetof(composite_2, two.d), "UB");
static_assert(offsetof(composite_1, e) == offsetof(composite_2, two.e), "UB");
You can define then as MACROs:
#define ENTITY_MEMBERS int x; int y
struct entity{
ENTITY_MEMBERS;
}
struct player {
ENTITY_MEMBERS;
char* name;
};
Actually this is how you mimic C++ single inheritance in C.
I want to know if it is possible to declare anonymous structs in ANSI C. The code I have is:
struct A
{
int x;
};
struct B
{
struct A;
int y;
};
When I compile it I get:
warning: declaration does not declare anything
I have read that the flag -fms-extensions does the trick, it however only works on windows systems as it produces:
warning: anonymous structs are a Microsoft extension [-Wmicrosoft]
Is there any ANSI equivalent extension that I can use?
A trick to get almost this feature in ANSI C is to use an appropriate macro:
struct A {
int x;
};
struct B {
struct A A_;
int y;
};
#define bx A_.x
Then you can simply do
struct B foo, *bar;
foo.bx;
bar->bx;
In C11 though, anonymous structures are supported and you can simply do
struct B {
struct {
int x;
};
int y;
}
but sadly not
struct A {
int x;
};
struct B
{
struct A;
int y;
};
As the anonymous structure has to be declared inside the structure it is anonymous to.
See this answer for more details on anonymous members in C11.
Its possible to declare anonymous struct and union. The ISO C11 added this feature and GCC allows it as an extension.
C11 section §6.7.2.1 para 13:
An unnamed member whose type specifier is a structure specifier with no tag is called an anonymous structure; an unnamed member whose type specifier is a union specifier with no tag is called an anonymous union. The members of an anonymous structure or union are considered to be members of the containing structure or union. This applies recursively if the containing structure or union is also anonymous.
19 The following illustrates anonymous structures and unions:
struct v {
union { // anonymous union
struct { int i, j; }; // anonymous structure
struct { long k, l; } w;
};
int m;
} v1;
v1.i = 2; // valid
v1.k = 3; // invalid: inner structure is not anonymous
v1.w.k = 5; // valid
Now b can be accessed by just using foo.b.
You want something like this i suppose:
struct B {
struct {
int x;
} A;
int y;
};
And you could do:
struct B b;
b.A.x = 5;
printf( "%d\n", b.A.x );
I have a struct that declared as follow:
struct a {
struct b {
int d;
} c;
};
How to declare a variable of b outside a? In C++ I can use a::b x;. But, in C it required to specifies struct keyword before struct name.
C has a flat layout; when you declare a struct within another struct, the former is just being put into the global namespace.
So, in your example, it is just struct b.
C does not have nested types. You can't write a::x b or anything that ressembles it. If you want to get rid of the struct keyword, that's another problem. Use typedefs. but it won't allow to nest types.
typedef struct b_t {
int d;
} b;
typedef struct {
b c;
} a;
b some_b;
a some_a;
int f() {
some_b.d=42;
some_a.c=some_b;
return 0;
}
.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Arrow operator (->) usage in C
As far as i know only C++ can use classes(obj->something) however i have seen this operator in numerous C applications.
And a small side-question. Usually one uses structures in C like this:
structname.somevariable
However i have seen them used like:
structname.something1.something2
Does it have something to do with the keyword union?
struct A
{
int b;
};
struct A *a;
a->b == (*a).b;
And no, that has nothing to do with unions. It's just getting a member of a member.
if you have a pointer to a struct object, like
struct P * p;
you access members with ->
p->member
if p is not a pointer, you access members with .
struct P p;
p.member
Any C book covers this :P
Operator -> is used to access a member of a struct through a pointer to that structure. The second part of your question is used when accessing nested structures, it is not restricted to the use of unions. For instance:
struct A {
int a;
};
struct B {
struct A baz;
};
int main()
{
struct A foo;
struct B bar;
(&foo)->a = 10;
bar.baz.a = 20;
return 0;
}
C++ classes are an extension of C structs, and an object is just a pointer to a struct. C++ didn't actually invent a whole lot of new stuff; it's called C ++ for a reason.
structs can be nested. Given
struct a {
int a_a;
int a_b;
}
struct b {
int b_a;
struct a b_b;
} a;
you can refer to a.b_b.a_b.
A union uses similar syntax to a struct, but all the members overlap each other. This is useful when you need to interpret a chunk of memory in multiple ways.
I strongly suggest picking up a C book.
x.y.z is used when you want to access a member of a struct that is itself a member of another struct.
typedef struct _POINT
{
int x;
int y;
} POINT;
typedef struct _RECT
{
POINT topLeft;
POINT bottomRight;
} RECT;
int main()
{
RECT r;
r.topLeft.x = 0;
t.topLeft.y = 0;
int width = r.bottomRight.x - r.topLeft.x;
}
I've written a small example in C.
#include <stdio.h>
#include <stdlib.h>
struct a_t {
int a;
/* ... */
};
struct b_t {
struct a_t a;
/* ... */
};
int main()
{
struct a_t a;
struct a_t *b;
b = malloc(sizeof(struct a_t));
a.a = 1;
b->a = 2;
printf("%d; %d;\n", a.a, b->a);
(&a)->a = 3;
(*b).a = 4;
printf("%d; %d;\n", a.a, b->a);
free(b);
struct b_t c;
c.a.a = 5;
printf("%d;\n", c.a.a);
return 0;
}
If I remember correctly, in C++
class IDENTIFIER {
// stuff, possibly including functions
};
is exactly the same as
struct IDENTIFIER {
private:
// stuff, possibly including functions
};
and
struct IDENTIFIER {
// stuff, possibly including functions
};
is exactly the same as
class IDENTIFIER {
public:
// stuff, possibly including functions
};
I've seen C structs declared several different ways before. Why is that and what, if anything, does each do different?
For example:
struct foo {
short a;
int b;
float c;
};
typedef struct {
short d;
int e;
float f;
} bar;
typedef struct _baz {
short a;
int b;
float c;
} baz;
int main (int argc, char const *argv[])
{
struct foo a;
bar b;
baz c;
return 0;
}
Well, the obvious difference is demonstrated in your main:
struct foo a;
bar b;
baz c;
The first declaration is of an un-typedefed struct and needs the struct keyword to use. The second is of a typedefed anonymous struct, and so we use the typedef name. The third combines both the first and the second: your example uses baz (which is conveniently short) but could just as easily use struct _baz to the same effect.
Update: larsmans' answer mentions a more common case where you have to use at least struct x { } to make a linked list. The second case wouldn't be possible here (unless you abandon sanity and use a void * instead) because the struct is anonymous, and the typedef doesn't happen until the struct is defined, giving you no way to make a (type-safe) pointer to the struct type itself. The first version works fine for this use, but the third is generally preferred in my experience. Give him some rep for that.
A more subtle difference is in namespace placement. In C, struct tags are placed in a separate namespace from other names, but typedef names aren't. So the following is legal:
struct test {
// contents
};
struct test *test() {
// contents
}
But the following is not, because it would be ambiguous what the name test is:
typedef struct {
// contents
} test;
test *test() {
// contents
}
typedef makes the name shorter (always a plus), but it puts it in the same namespace as your variables and functions. Usually this isn't an issue, but it is a subtle difference beyond the simple shortening.
It's largely a matter of personal preference. I like to give new types a name starting with a capital letter and omit the struct, so I usually write typedef struct { ... } Foo. That means I cannot then write struct Foo.
The exception is when a struct contains a pointer to its own type, e.g.
typedef struct Node {
// ...
struct Node *next;
} Node;
In this case you need to also declare the struct Node type, since the typedef is not in scope within the struct definition. Note that both names may be the same (I'm not sure where the underscore convention originated, but I guess older C compilers couldn't handle typedef struct X X;).
All your uses are syntactically correct. I prefer the following usage
/* forward declare all structs and typedefs */
typedef struct foo foo;
.
.
/* declare the struct itself */
struct foo {
short a;
int b;
foo* next;
};
Observe that this easily allows to use the typedef already inside the declaration of the struct itself, and that even for struct that reference each other mutually.
The confusion comes about because some of the declarations are in fact declaring up to three C constructs. You need to keep in mind the difference between:
A typedef declaration,
A struct definition, and
A struct declaration.
They are all very different C constructs. They all do different things; but you can combine them into the one compound construct, if you want to.
Let's look at each declaration in turn.
struct foo {
short a;
int b;
float c;
};
Here we are using the most basic struct definition syntax. We are defining a C type and give it the name foo in the tag namespace. It can later be used to declare variables of that type using the following syntax:
struct foo myFoo; // Declare a struct variable of type foo.
This next declaration gives the type another name (alias) in the global namespace. Let's break it down into its components using the previous basic declaration.
typedef foo bar; // Declare bar as a variable type, the alias of foo.
bar myBar; // No need for the "struct" keyword
Now just replace "foo" with the the struct's definition and voila!
typedef struct {
short d;
int e;
float f;
} bar;
typedef struct _baz {
short a;
int b;
float c;
} baz;
The above syntax is equivalent to the following sequence of declarations.
struct _baz {
short a;
int b;
float c;
}
typedef _baz baz; // Declare baz as an alias for _baz.
baz myBaz; // Is the same as: struct _baz myBaz;