I'm really new to C programming and I'm still trying to understand the concept of using pointers and using typedef structs.
I have this code snippet below that I need to use in a program:
typedef struct
{
char* firstName;
char* lastName;
int id;
float mark;
}* pStudentRecord;
I'm not exactly sure what this does - to me it seems similar as using interfaces in Objective-C, but I don't think that's the case.
And then I have this line
pStudentRecord* g_ppRecords;
I basically need to add several pStudentRecord to g_ppRecords based on a number. I understand how to create and allocate memory for an object of type pStudentRecord, but I'm not sure how to actually add multiple objects to g_ppRecords.
defines a pointer to the struct described within the curly bracers, here is a simpler example
typedef struct {
int x;
int y;
}Point,* pPoint;
int main(void) {
Point point = {4,5};
pPoint point_ptr = &point;
printf("%d - %d\n",point.x,point_ptr->x);
pPoint second_point_ptr = malloc(sizeof(Point));
second_point_ptr->x = 5;
free(second_point_ptr);
}
The first declares an unnamed struct, and a type pStudentRecord that is a pointer to it. The second declares g_ppRecords to be a pointer to a pStudentRecord. In other words, a pointer to a pointer to a struct.
It's probably easier to think of the second as an "array of pointers". As such, g_ppRecords[0] may point to a pStudentRecord and g_ppRecords[1] to another one. (Which, in turn, point to a record struct.)
In order to add to it, you will need to know how it stores the pointers, that is, how one might tell how many pointers are stored in it. There either is a size somewhere, which for size N, means at least N * sizeof(pStudentRecord*) of memory is allocated, and g_ppRecords[0] through g_ppRecords[N-1] hold the N items. Or, it's NULL terminated, which for size N, means at least (N+1) * sizeof(pStudentRecord*) of memory is allocated and g_ppRecords[0] through g_ppRecords[N-1] hold the N items, and g_ppRecords[N] holds NULL, marking the end of the string.
After this, it should be straightforward to create or add to a g_ppRecords.
A struct is a compound data type, meaning that it's a variable which contains other variables. You're familiar with Objective C, so you might think of it as being a tiny bit like a 'data only' class; that is, a class with no methods. It's a way to store related information together that you can pass around as a single unit.
Typedef is a way for you to name your own data types as synonyms for the built-in types in C. It makes code more readable and allows the compiler to catch more errors (you're effectively teaching the compiler more about your program's intent.) The classic example is
typedef int BOOL;
(There's no built-in BOOL type in older ANSI C.)
This means you can now do things like:
BOOL state = 1;
and declare functions that take BOOL parameters, then have the compiler make sure you're passing BOOLs even though they're really just ints:
void flipSwitch(BOOL isOn); /* function declaration */
...
int value = 0;
BOOL boolValue = 1;
flipSwitch(value); /* Compiler will error here */
flipSwitch(boolValue); /* But this is OK */
So your typedef above is creating a synonym for a student record struct, so you can pass around student records without having to call them struct StudentRecord every time. It makes for cleaner and more readable code. Except that there's more to it here, in your example. What I've just described is:
typedef struct {
char * firstName;
char * lastName;
int id;
float mark;
} StudentRecord;
You can now do things like:
StudentRecord aStudent = { "Angus\n", "Young\n", 1, 4.0 };
or
void writeToParents(StudentRecord student) {
...
}
But you've got a * after the typedef. That's because you want to typedef a data type which holds a pointer to a StudentRecord, not typedef the StudentRecord itself. Eh? Read on...
You need this pointer to StudentRecord because if you want to pass StudentRecords around and be able to modify their member variables, you need to pass around pointers to them, not the variables themselves. typedefs are great for this because, again, the compiler can catch subtle errors. Above we made writeToParents which just reads the contents of the StudentRecord. Say we want to change their grade; we can't set up a function with a simple StudentRecord parameter because we can't change the members directly. So, we need a pointer:
void changeGrade(StudentRecord *student, float newGrade) {
student->mark = newGrade;
}
Easy to see that you might miss the *, so instead, typedef a pointer type for StudentRecord and the compiler will help:
typedef struct { /* as above */ } *PStudentRecord;
Now:
void changeGrade(PStudentRecord student, float newGrade) {
student->mark = newGrade;
}
It's more common to declare both at the same time:
typedef struct {
/* Members */
} StudentRecord, *PStudentRecord;
This gives you both the plain struct typedef and a pointer typedef too.
What's a pointer, then? A variable which holds the address in memory of another variable. Sounds simple; it is, on the face of it, but it gets very subtle and involved very quickly. Try this tutorial
This defines the name of a pointer to the structure but not a name for the structure itself.
Try changing to:
typedef struct
{
char* firstName;
char* lastName;
int id;
float mark;
} StudentRecord;
StudentRecord foo;
StudentRecord *pfoo = &foo;
Related
#define LENGTH 6
typedef char data_t[LENGTH];
struct foo {
const data_t data;
...
}
...
void bar(data_t data) {
printf("%.6s\n", data);
struct foo myfoo = {*data};
printf("%.6s\n", foo.data);
}
I'm trying to have this struct which holds directly the data I'm interested in, sizeof(foo) == 6+the rest, not sizeof(foo) == sizeof(void*)+the rest. However I can't find a way to initialize a struct of type foo with a data_t. I think maybe I could remove the const modifier from the field and use memcpy but I like the extra safety and clarity.
I don't get any compile errors but when I run the code I get
123456
1??
so the copy didn't work properly I think.
This is for an arduino (or similar device) so I'm trying to keep it to very portable code.
Is it just not possible ?
EDIT: removing the const modifier on the data_t field doesn't seem to help.
It is possible to do this, for some cost >=0.
typedef struct
{
char c[LENGTH];
} data_t; // this struct is freely copyable
struct foo
{
const data_t data; // but this data member is not
int what;
};
void foo (char* x) {
data_t d; // declare freely copyable struct instance
memcpy(d.c, x, sizeof(d.c)); // memcpy it
struct foo foo = { d, 42 }; // initialise struct instance with const member
...
};
Some compilers (e.g. clang) are even able to optimise away the redundant copying (from x to d.c and then from d to foo.data ⇒ from x straight to foo.data). Others (gcc I'm looking at you) don't seem to be able to achieve this.
If you pass around pointers to data_t rather than straight char pointers, you won't need this additional memcpy step. OTOH in order to access the char array inside foo you need another level of member access (.data.c instead of just .data; this has no runtime cost though).
It's impossible to do it in a standard compliant way.
Due to its being const, const char data[6]; must be initialized to be usable, and it may only be initialized statically (static objects with no initializer get automatically zeroed), with a string literal, or with a brace-enclosed initializer list. You cannot initialize it with a pointer or another array.
If I were you, I would get rid of the const, document that .data shouldn't be changed post-initialization, and then use memcpy to initialize it.
(const on struct members doesn't work very well in my opinion. It effectively prevents you from being able to have initializer functions, and while C++ gets around the problem a little bit by having special language support for its constructor functions, the problem still remains if the const members are arrays).
I am using MinGW on Windows. I am building linked list and I am confused with this.
#include <stdio.h>
#include <stdlib.h>
typedef struct Data
{
int x;
int y;
struct BlaBla * next; /*compiles with no problem*/
}List;
int main(void)
{
List item;
List * head;
head = NULL;
return 0;
}
I now that struct can't have struct variable(object, instance of that struct), but can have pointer of that struct type. Didn't know that pointer can be pointer of unexisting type. struct BlaBla * next;(not for linked list, it must be struct Data * next but mean general talking)
Yes, you can, because then the compiler, upon encountering the unknown type name for the first time, assumes that there's somehwere a struct type definition with this name. Then it will forward-declare the struct name for you, let you use it as a pointer, but you can't dereference it nor can you do pointer arithmetic on it (since it's an incomplete type).
The compiler will accept code such as your example:
typedef struct Data
{
int x;
int y;
struct BlaBla * next; /*compiles with no problem*/
}List;
This is okay because the size of pointers is known to the compiler, and the compiler is assuming that the struct will be defined before it is dereferenced.
Because the compiler acts this way, it's possible to do this:
typedef struct Data
{
int x;
int y;
struct Data * next; /* points to itself */
} List;
However, if you were to include the struct inline, like this:
typedef struct Data
{
int x;
int y;
struct BlaBla blaStruct; /* Not a pointer. Won't compile. */
}List;
The compiler can't work out how big struct Data is because it doesn't know how big struct BlaBla is. To get this to compile, you need to include the definition of struct BlaBla.
Note that, as soon as you need to access the members of struct BlaBla, you will need to include the header file that defines it.
It depends on what you mean by "unexisting". If you haven't even declared BlaBla, you'll get an error.
If you've declared it but not yet defined it, that will work fine. You're allowed to have pointers to incomplete types.
In fact, that's the normal way of doing opaque pointers in C.
So, you might think that this is invalid because there's no declaration of struct BlaBla in scope:
typedef struct Data {
struct BlaBla *next; // What the ??
} List;
However, it's actually okay since it's both declaring struct BlaBla and defining next at the same time.
Of course, since definition implies declaration, this is also okay:
struct BlaBla { int xyzzy; };
typedef struct Data {
struct BlaBla *next; // What the ??
} List;
In order to declare a variable or field of a given type, pass one as a parameter, or copy one to another of the same type, the compiler has to know how many bytes the variable or field occupies, what alignment requirements it has (if any), and what other pointer types it's compatible with, but that's all the compiler needs to know about it. In all common dialects of C, a pointer to any structure will always be the same size and require the same alignment, regardless of the size of the structure to which it points or what that structure may contain, and pointers to any structure type are only compatible with other pointers to the same structure type.
Consequently, code which doesn't need to do anything with pointers to a structure except allocate space to hold the pointers themselves [as opposed to the structures at which they point], pass them as parameters, or copy them to other pointers, doesn't need to know anything about the structure type to which they point beyond its unique name. Code which needs to allocate space for a structure (as opposed to a pointer to one) or access any of its members must know more about its type, but code which doesn't do those things doesn't need such information.
I would like to simulate the object oriented programming, so in C++, let's consider the following C code:
typedef struct tAnimal{
char * name;
int age;
}tAnimal;
typedef struct tAnimal2{
char * name;
int age;
float size;
}tAnimal2;
In C++ you can create a table of different objects which are inherited from the same class.
I would like to do the same in C, let's consider the following code:
tAnimal ** tab;
tab = malloc(sizeof(tAnimal*)*2);
tab[0] = malloc(sizeof(tAnimal));
tab[1] = malloc(sizeof(tAnimal2));
Notice that the allocation works because malloc returns a void pointer, and C does not require casting. But I still have no access to the size field, because the type of tab elements is tAnimal after all.
Is there anyway to fix this?, I would like to stay away from void ** pointers.
In C it's common to use a structure with a type-flag, and a union of the data:
typedef enum
{
Animal1,
Animal2
} AnimalType;
struct Animal
{
AnimalType type;
union
{
tAnimal animal;
tAnimal2 animal2;
};
};
Now you can create an array of the Animal structure.
If you want to access the size field you have to cast your pointer to tAnimal2. Note, the same would be true for C++.
You can kind of simulate inheritance by embedding the first struct at the beginning of the second:
struct tAnimal{
char * name;
int age;
};
struct tAnimal2{
struct tAnimal parent;
float size;
};
In order to access the size field in tab[1] you can cast the pointer to a tAnimal2 pointer.
tAnimal2* panimal2 = (tAnimal2*) tab[1];
panimal2->size = 1.0;
However this practice is prone to data corruption since you will need a method to ensure that the element of the table you cast to tAnimal2 is indeed an instance of tAnimal2. You could use an additional type field as Joachim Pileborg suggests to check the type of the object.
Often stacks in C are dependent upon datatype used to declare them. For example,
int arr[5]; //creates an integer array of size 5 for stack use
char arr[5]; //creates a character array of size 5 for stack use
are both limited to holding integer and character datatypes respectively and presumes that the programmer knows what data is generated during the runtime. What if I want a stack which can hold any datatype?
I initially thought of implementing it as a union, but the approach is not only difficult but also flawed. Any other suggestions?
I would use a structure like this:
struct THolder
{
int dataType; // this is a value representing the type
void *val; // this is the value
};
Then use an array of THolder to store your values.
This is really just a variant of Pablo Santa Cruz' answer, but I think it looks neater:
typedef enum { integer, real, other } type_t;
typedef struct {
type_t type;
union {
int normal_int; /* valid when type == integer */
double large_float; /* valid when type == real */
void * other; /* valid when type == other */
} content;
} stack_data_t;
You still need to use some way to explicitly set the type of data stored in each element, there is no easy way around that.
You could look into preprocessor magic relying on the compiler-dependent typeof keyword to do that automagically, but that will probably not do anything but ruin the portability.
Some people have suggested a void* member. In addition to that solution I'd like to offer an alternative (assuming your stack is a linked list of heap-allocated structures):
struct stack_node
{
struct stack_node *next;
char data[];
};
The data[] is a C99 construct. data must be the last member; this takes advantage of the fact that we can stuff arbitrary quantities after the address of the struct. If you're using non-C99 compiler you might have to do some sketchy trick like declare it as data[0].
Then you can do something like this:
struct stack_node*
allocate_stack_node(size_t extra_size)
{
return malloc(sizeof(struct stack_node) + extra_size);
}
/* In some other function... */
struct stack_node *ptr = allocate_stack_node(sizeof(int));
int *p = (int*)ptr->data;
If this looks ugly and hacky, it is... But the advantage here is that you still get the generic goodness without introducing more indirection (thus slightly quicker access times for ptr->data than if it were void* pointing to a different location from the structure.)
Update: I'd also like to point out that the code sample I give may have problems if your machine happens to have different alignment requirements for int than char. This is meant as an illustrative example; YMMV.
You could use macros and a "container" type to reduce "type" from being per-element, to whole-container. (C99 code below)
#define GENERIC_STACK(name, type, typeid, elements) \
struct name##_stack { \
unsigned int TypeID; \
type Data[elements]; \
} name = { .TypeID = typeid }
Of course, your "TypeID" would have to allow every possible agreed-upon type you expect; might be a problem if you intend to use whole structs or other user-defined types.
I realize having a uniquely named struct type for every variable is odd and probably not useful... oops.
I created an library that works for any data type:
List new_list(int,int);
creates new list eg:
List list=new_list(TYPE_INT,sizeof(int));
//This will create an list of integers
Error append(List*,void*);
appends an element to the list. *Append accpts two pointers as an argument, if you want to store pointer to the list don't pass the pointer by pointer
eg:
//using the int list from above
int a=5;
Error err;
err=append(&list,&a)
//for an list of pointers
List listptr=new_list(TYPE_CUSTOM,sizeof(int*));
int num=7;
int *ptr=#
append(&listptr,ptr);
//for list of structs
struct Foo
{
int num;
float *ptr;
};
List list=new_list(TYPE_CUSTOM,sizeof(struct Foo));
struct Foo x;
x.num=9;
x.ptr=NULL;
append(&list,&x);
Error get(List*,int);
Gets data at index specified. When called list's current poiter will point to the data.
eg:
List list=new_list(TYPE_INT,sizeof(int));
int i;
for(i=1;i<=10;i++)
append(&list,&i);
//This will print the element at index 2
get(&list,2);
printf("%d",*(int*)list.current);
Error pop(List*,int);
Pops and element from the specified index
eg:
List list=new_list(TYPE_INT,sizeof(int));
int i;
for(i=1;i<=10;i++)
append(&list,&i);
//element in the index 2 will be deleted,
//the current pointer will point to a location that has a copy of the data
pop(&list,2);
printf("%d",*(int*)list.current);
//To use the list as stack, pop at index list.len-1
pop(&list,list.len-1);
//To use the list as queue, pop at index 0
pop(&list,0);
Error merge(List ,List);
Merges two list of same type. If types are different will return a error message in the Error object it returns;
eg:
//Merge two elements of type int
//List 2 will come after list 1
Error err;
err=merge(&list1,&list2);
Iterator get_iterator(List*);
Get an iterator to an list. when initialized will have a pointer to the first element of the list.
eg:
Iterator ite=get_iterator(&list);
Error next(Iterator*);
Get the next element of the list.
eg:
//How to iterate an list of integers
Iterator itr;
for(itr=get_iterator(&list); ite.content!=NULL; next(ite))
printf("%d",*(int*)ite.content);
https://github.com/malayh/C-List
Sometimes I see code like this (I hope I remember it correctly):
typedef struct st {
int a; char b;
} *stp;
While the usual pattern that I familiar with, is:
typedef struct st {
int a; char b;
} st;
So what's the advantage in the first code example?
You probably mean this:
typedef struct ST {
/* fields omitted */
} *STP;
The asterisk is at the end of the statement. This simply means "define the type STP to be a pointer to a struct of this type". The struct tag (ST) is not needed, it's only useful if you want to be able to refer to the struct type by itself, later on.
You could also have both, like so:
typedef struct {
/* fields omitted */
} ST, *STP;
This would make it possible to use ST to refer to the struct type itself, and STP for pointers to ST.
Personally I find it a very bad practice to include the asterisk in typedefs, since it tries to encode something (the fact that the type is a pointer) into the name of the type, when C already provides its own mechanism (the asterisk) to show this. It makes it very confusing and breaks the symmetry of the asterisk, which appears both in declaration and use of pointers.
It's a habit that stems from the time when typedef names and struct tagnames were in the same namespace. See http://blogs.msdn.com/oldnewthing/archive/2008/03/26/8336829.aspx
I think you are talking about :
typedef struct{
int a;
char b;
} object, *objectPointer;
This means that (new) type objectPointer is a pointer to struct (object) defined above. Its easy to declare pointers to object struct this way. For instance,
objectPointer A = (objectPointer)malloc(sizeof(object));
A->a = 2;
Now, A is a pointer to struct object and you can access its variables as described above.
In case, objectPointer was not defined,
struct object *A = (struct object *)malloc(sizeof(object));
A->a = 2;
So, I guess objectPointer is more intuitive and easy to use.
I hope that the first code would say a compiler error ,
I see no good reason for the typedef name be different from the tag name.
Now, the reason for which the tag name needs to be typedefed if you don't want to use
struct tag v;
but
tag v;
is probably an historical one. For as long as I remember, C had typedef but I don't know if it was true when struct have been introduced (handling of typedef is a nuisance in the C grammar). In the old code I've seen, using typedef for struct isn't done, and there are things like unix
struct stat;
int stat(const char*, struct stat*);
which would break with an automatic typedef. One those are introduced, changing is quite difficult (yes, C++ has automatic typedef but C++ has special wording to handle that case of overloading and it would be yet another complication).