I am trying to solve a little problem from my university, but i get into trouble completing it.
The task is to write a code defining all the data structures and create a link between them using pointers and mallocs.
ASSIGNMENT :
There are 3 blocks.
1) is a static block with 2 fields, 1st field is an integer, and a second field points to another block.
2) a block with 2 fields. 1st field points to the first block and 2nd field points to the 3rd block.
3) a block with 2 fields. 1st field is an integer and 2nd field points to the first block.
I have started creating the code and making the links but i keep getting errors which i cant seem to find thier logic.
struct A;
struct B;
struct C;
typedef struct {
int element;
struct A * pointer3;
}C;
typedef struct {
struct A * pointer 1;
struct C * pointer 2;
}B;
typedef struct {
int element;
struct B * pointer 0;
}A;
int main (){
A a;
a.pointer0 = (B*)malloc(sizeof(B));
I have reached farther then that, but the problem i get is in the last line.
Gives me the error warning: assignment from incompatible pointer type
Question: What does struct A look like? Answer: Your compiler has no idea because you never defined it.
You're assuming forward-declaring a structure type, then formally declaring a typedef alias to a structure where the alias name (but not tag) is the same as the structure tag (with no name) will resolve. It won't. You have to do that yourself.
Start with simple. Get rid of the type aliases entirely. And get rid of any unnecessary forward decls as well (B and C) What would it look like?
#include <stdio.h>
#include <stdlib.h>
struct A; // fwd decl
struct C
{
int element;
struct A *pointer3; // uses fwd decl
};
struct B
{
struct A *pointer1; // uses fwd decl
struct C *pointer2; // uses prior def
};
struct A // resolves fwd decl
{
int element;
struct B *pointer0; // uses prior def
};
int main()
{
struct A a;
a.pointer0 = malloc(sizeof(struct B));
}
That works (in the most optimistic definition of the term). If you want to use typedef aliases, you can add them in now. For example, here is all three structures both tagged and aliased.
#include <stdio.h>
#include <stdlib.h>
struct A;
typedef struct A A;
typedef struct C
{
int element;
A *pointer3;
} C;
typedef struct B
{
A *pointer1;
C *pointer2;
} B;
struct A
{
int element;
B *pointer0;
};
int main()
{
A a;
a.pointer0 = malloc(sizeof(B));
}
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.
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 */
I am a beginner in C programming and I know the difference between struct type declaration and typedef struct declaration. I came across to know an answer saying that if we define a struct like:
typedef struct {
some members;
} struct_name;
Then it will be like providing an alias to an anonymous struct (as it is not having a tag name). So it can't be used for forward declaration. I don't know what the forward declaration means.
Also, I wanted to know that for the following code:
typedef struct NAME {
some members;
} struct_alias;
Is there any difference between NAME and struct_alias? Or are both equal as
struct_alias is an alias of struct NAME ?
Furthermore, can we declare a variable of type struct NAME like these:
struct_alias variable1;
and/or like:
struct NAME variable2;
or like:
NAME variable3;
struct forward declarations can be useful when you need to have looping struct declarations. Example:
struct a {
struct b * b_pointer;
int c;
};
struct b {
struct a * a_pointer;
void * d;
};
When struct a is declared it doesn't know the specs of struct b yet, but you can forward reference it.
When you typedef an anonymous struct then the compiler won't allow you to use it's name before the typedef.
This is illegal:
struct a {
b * b_pointer;
int c;
};
typedef struct {
struct a * a_pointer;
void * d;
} b;
// struct b was never declared or defined
This though is legal:
struct a {
struct b * b_pointer;
int c;
};
typedef struct b {
struct a * a_pointer;
void * d;
} b;
// struct b is defined and has an alias type called b
So is this:
typedef struct b b;
// the type b referes to a yet undefined type struct b
struct a {
b * struct_b_pointer;
int c;
};
struct b {
struct a * a_pointer;
void * d;
};
And this (only in C, illegal in C++):
typedef int b;
struct a {
struct b * struct_b_pointer;
b b_integer_type;
int c;
};
struct b {
struct a * a_pointer;
void * d;
};
// struct b and b are two different types all together. Note: this is not allowed in C++
Forward declaration is a promise to define something that you make to a compiler at the point where the definition cannot be made. The compiler can use your word to interpret other declarations that it would not be able to interpret otherwise.
A common example is a struct designed to be a node in a linked list: you need to put a pointer to a node into the struct, but the compiler would not let you do it without either a forward declaration or a tag:
// Forward declaration
struct element;
typedef struct {
int value;
// Use of the forward declaration
struct element *next;
} element; // Complete definition
and so it cant be used for forward declaration
I think that author's point was that giving your struct a tag would be equivalent to a forward declaration:
typedef struct element {
int value;
// No need for a forward declaration here
struct element *next;
} element;
Forward declaration is a declaration preceeding an actual definition, usually for the purpose of being able to reference the declared type when the definition is not available. Of course, not everything may be done with the declared-not-defined structure, but in certain context it is possible to use it. Such type is called incomplete, and there are a number of restrictions on its usage. For example:
struct X; // forward declaration
void f(struct X*) { } // usage of the declared, undefined structure
// void f(struct X) { } // ILLEGAL
// struct X x; // ILLEGAL
// int n =sizeof(struct X); // ILLEGAL
// later, or somewhere else altogether
struct X { /* ... */ };
This can be useful e.g. to break circular dependencies, or cut down the compilation time, as the definitions are usually significantly larger, and so more resources are required to parse it.
In your example, struct NAME and struct_alias are indeed equivalent.
struct_alias variable1;
struct NAME variable2;
are correct;
NAME variable3;
is not, as in C the struct keyword is required.
struct_alias and struct NAME are same ,struct_alias is an alias to struct NAME
These both are same and allowed
struct_alias variable1;
struct NAME variable1;
this is illegal
NAME variable3;
See this article on Forward declaration
As others stated before, a forward declaration in C/C++ is the declaration of something with the actual definition unavailable. Its a declaration telling the compiler "there is a data type ABC".
Lets pretend this is a header for some key/value store my_dict.h :
...
struct my_dict_t;
struct my_dict_t* create();
char* get_value(const struct my_dict_t* dict, const char* name);
char* insert(struct my_dict_t* dict, const char* name, char* value);
void destroy(struct my_dict_t* dict);
...
You dont know anything about my_dict_t, but actually, for using the store
you dont need to know:
#include "my_dict.h"
...
struct my_dict_t* dict = create();
if(0 != insert(dict, "AnEntry", strdup("AValue"))) {
...
}
...
The reason for this is: You are only using POINTERS to the data structure.
POINTERS are just numbers, and for dealing with them you dont need to know what they are pointing at.
This will only matter if you try to actually access them, like
struct my_dict_t* dict = create();
printf("%s\n", dict->value); /* Impossible if only a forward decl is available */
So, for implementing the functions, you require an actual definition of my_struct_t.
You might do this in the source file my_dict.c like so:
#include "my_dict.h"
struct my_dict_t {
char* value;
const char* name;
struct my_dict_t* next;
}
struct my_dict_t* create() {
return calloc(1, sizeof(struct my_dict_t));
}
This is handy for several situations, like
For resolving circular type dependencies, like Sergei L. explained.
For encapsulation, like in the example above.
So the question that remains is: Why cant we just omit the forward declaration at all when using the functions above? In the end, it would suffice for the compiler to know that all dict are pointers.
However, the compiler does perform type checks:
It needs to verify that you don't do something like
...
int i = 12;
char* value = get_value(&i, "MyName");
...
It does not need to know how my_dict_t looks like, but it needs to know that &i is not the type of pointer get_value() expects.
This produces an incompatibility warning:
#include <stdlib.h>
#include <stdio.h>
typedef struct
{
int key;
int data;
struct htData_* next;
struct htData_* prev;
}htData_;
typedef struct
{
int num_entries;
struct htData_** entries;
}ht_;
ht_* new_ht(int num_entries);
int ht_add(ht_* ht_p, int key, int data);
int main()
{
int num_entries = 20;
//crate a hash table and corresponding reference
ht_* ht_p = new_ht(num_entries);
//add data to the hash table
int key = 1305;
ht_add(ht_p,key%num_entries,20);
return 0;
}
ht_* new_ht(int num_entries)
{
ht_ *ht_p;
ht_ ht;
ht.num_entries = num_entries;
ht_p = &ht;
//create an array of htData
htData_ *htDataArray;
htDataArray = (htData_*) malloc(num_entries * sizeof(htData_));
//point to the pointer that points to the first element in the array
ht.entries = &htDataArray; // WARNING HERE!!!!!!!!!!!!!!!!
return ht_p;
}
I'm trying to copy the **ptr to the struct containing a **ptr.
Update: My simplified code was not accurate so I've posted the actual code.
The problem is that struct htData_ and htData_ are not the same thing! As far as the compiler is concerned, struct htData_ doesn't exist—it's an incomplete type. htData_, on the other hand, is a typedef for an anonymous structure. For a more detailed analysis, see Difference between struct and typedef struct in C++.
So, you're getting a warning because ht.entries is declared as the type struct htData_**, but the right-hand side of that assignment has type <anonymous struct>**. To fix this, you need to define struct htData_:
typedef struct htData_
{
...
} htData_;
This line is not proper:
htData_ array[20] = htDataArray;
You cannot assign a pointer to an array.
In your edited code, here is the problematic line:
//point to the pointer that points to the first element in the array
ht.entries = &htDataArray;
Actually, syntactically it's correct, so it should not give warning. But you are doing wrong stuff here. If you want ht.entries pointing to the first element of array than you need to declare it as,
htData_* entries; // 'struct' keyword not needed ahead of declaration
and assign it as,
ht.entries = &htDataArray[0];
How can I have a pointer to the next struct in the definition of this struct:
typedef struct A {
int a;
int b;
A* next;
} A;
this is how I first wrote it but it does not work.
You can define the typedef and forward declare the struct first in one statement, and then define the struct in a subsequent definition.
typedef struct A A;
struct A
{
int a;
int b;
A* next;
};
Edit: As others have mentioned, without the forward declaration the struct name is still valid inside the struct definition (i.e. you can used struct A), but the typedef is not available until after the typedef definition is complete (so using just A wouldn't be valid). This may not matter too much with just one pointer member, but if you have a complex data structure with lots of self-type pointers, may be less wieldy.
In addition to the first answer, without a typedef and forward declaration, this should be fine too.
struct A
{
int a;
int b;
struct A *next;
};
You are missing the struct before the A*
typedef struct A {
int a;
int b;
struct A* next;
} A;
You can go without forward declaration:
struct A {
int a;
int b;
struct A *next;
};
Please, you're in C, not C++.
If you really must typedef a struct (and most programmers that I work with would not¹), do this:
typedef struct _A {
int a;
int b;
struct _A *next;
} A;
to clearly differentiate between _A (in the struct namespace) and A (in the type namespace).
¹typedef hides the size and storage of the type it points to ― the argument (and I agree) is that in a low-level language like C, trying to hide anything is harmful and counterproductive. Get used to typing struct A whenever you mean struct A.
typedef struct {
values
} NAME;
This is shorter way to typedef a struct i think its the easiest notation, just don't put the name infront but behind.
you can then call it like
NAME n;
NAME *n; // if you'd like a ptr to it.
Anything wrong with this approach?