Example:
struct dummy
{
int var;
};
Why structures like this are used? Mostly I have seen them in some header files.
The atomic_t type is also defined like this. Can't it be defined simply using:
typedef int atomic_t;
It's more extensible.
Assume that in the future, you realize that struct dummy should contain a name field, then you can change the definition of it to:
struct dummy
{
int var;
char name[30];
};
without changing much of your application code.
Besides extensibility, this idiom also makes it syntactically impossible to do normal arithmetic on types whose meaning is such that that doesn’t make sense semantically.
E.g.:
typedef uint32_t myObject;
myObject x, y;
...
y = x + 3; // meaningless, but doesn’t produce an error.
// may later cause runtime failure.
v.s.
typedef struct { uint32_t var; } myObject;
myObject x, y;
...
y = x + 3; // syntax error.
This may seem contrived, but it is occasionally very useful.
One other use is to pass entire arrays to functions.
struct s {
int a[3];
};
void f1(int a[]) // this evaluates to a pointer, same as void f1(int *a)
{
printf("%d\n", sizeof(a));
}
void f2(struct s *obj)
{
printf("%d\n", sizeof(obj->a));
}
int main(int argc, char **argv)
{
int a[3] = {1, 2, 3};
struct s obj;
obj.a[0] = 1;
obj.a[1] = 2;
obj.a[2] = 3;
f1(a);
f2(&obj);
return 0;
}
// output
// 8
// 12
Not all things that are representable in 32 bits should be treated as numbers. Even things that have a numeric value may have semantics which would suggest that they need special treatment. Suppose, for example, that a processor has an "atomic increment" instruction but it's slower than a "normal" increment instruction. If one wants to atomically increment fnord in one place and decrement it in another, one could use:
volatile int fnord;
...
atomic_inc(&fnord);
...
atomic_dec(&fnord);
A problem with that, however, is that if one of the places that is supposed to increment fnord happens to use fnord++ rather than atomic_inc(&fnord);, the compiler will perfectly happily generate a "normal" increment instruction, and the code may work most of the time, but it could fail in hard-to-track down fashion.
Replacing the int with a structure (and defining atomic_inc inline functions to work with it) would prevent erroneous code like fnord++; from compiling. It wouldn't guard against fnord.var++; but would give a programmer a chance to examine the structure and see what the right way to increment it would be.
Mostly it is to keep compatibility, as maybe earlier the structure had additional elements.
Or as it could be intended to add other elements later.
(or even an itnernal version of the structure has more than just one member (what I realy could imagine for atomic_t-type.)
Related
Given I'll return a large struct in a function like here:
#include <stdio.h>
// this is a large struct
struct my_struct {
int x[64];
int y[64];
int z[64];
};
struct my_struct get_my_struct_from_file(const char *filename) {
int tmp1, tmp2; // some tmp. variables
struct my_struct u;
// ... load values from filename ...
return u;
}
int main() {
struct my_struct res = get_my_struct_from_file("tmp.txt"); // <-- here
printf("x[0] = %d\n", res.x[0]);
// ... print all values ...
}
At the place marked by here, do I have to assume that this large struct is copied or is it likely that the compiler does something to avoid this?
Thank you
… do I have to assume that this large struct is copied…
No, of course you do not have to make that assumption. Nobody requires you to make that assumption, and it would be unwise to adopt the statement as an assumption rather than deriving it from known information, such as compiler documentation or inspection of the generated assembly code.
In the specific code you show, it is likely good compilers will optimize so that the structure is not copied. (Testing with Apple Clang 11 confirms it does this optimization.) But that is likely overly simplified code. If a call to get_my_struct_from_file appears in a translation unit separate from its definition, the compiler will not know what get_my_struct_from_file is accessing. If the destination object, res in this example, has had its address previously passed to some other routine in some other translation unit, then the compiler cannot know that other routine did not stash the address somewhere and that get_my_struct_from_file is not using it. So the compiler would have to treat the structure returned by get_my_struct_from_file and the structure the return value is being assigned to as separate; it could not coalesce them to avoid the copy.
To ensure the compiler does what you want, simply tell it what you want it to do. Write the code so that the function puts the results directly in the structure you want to put it in:
void get_my_struct_from_file(struct my_struct *result, const char *filename)
{
…
}
...
get_my_struct_from_file(&res, "tmp.txt");
At the place marked by here, do I have to assume that this large struct is copied or is it likely that the compiler does something to avoid this?
Semantically, the structure is copied from the function's local variable to the caller's variable. These are distinct objects, and just like objects of other types, setting one structure equal to another requires copying from the representation of one to the representation of the other.
The only way to avoid a copy would be for the compiler to treat the local variable as an alias for the caller's structure, but that would be wrong in the general case. Such aliasing can easily produce observably different behavior than would occur without.
It is possible that in some specific cases, the compiler can indeed avoid the copy, but if you want to ensure that no copying happens then you should set up the wanted aliasing explicitly:
void get_my_struct_from_file(const char *filename, struct my_struct *u) {
int tmp1, tmp2; // some tmp. variables
// ... load values from filename into *u
}
int main() {
struct my_struct res = { 0 };
get_my_struct_from_file("tmp.txt", &res);
printf("x[0] = %d\n", res.x[0]);
// ... print all values ...
}
I have the following c code:
struct {
short s;
int n;
} variableName;
I want to write a function to capture this variable like so
void func(MyStruct* var){
//do stuff
}
func(&variableName);
I would like to do this without providing a definition for the struct. Is there a way to capture variableName?
No, you can't pass an "anonymous" struct into a function in C. You could of course define your function to accept the arguments individually:
void func(short s, int n) { ... }
Or you can define the MyStruct structure in a place that both the function and the calling code has visibility to. Note that the whole struct is passed by value (copy) when you do that, which may be the behavior you want here (or may not be).
You may be looking for something more like a "dictionary" or "associative array" or "hash" type that many other languages provide, with arbitrary key value pairs in it. Pure C does not have a facility for this; the compiler wants to know the layout of a structure in advance.
(I'm not sure if you might be asking about a slightly more esoteric idea, which is hiding the composition of a structure and passing around an "opaque handle" out of and into an API. There are ways to structure that in C, but please say so if that's what you're talking about.)
Completely overlooked "I would like to do this without providing a definition for the struct. Is there a way to capture variableName?" in the OP, unless it was edited after. The question makes less sense now, but heres how you could normally pass a struct to a function for future readers.
#include <stdio.h>
struct StructName{
short s;
int n;
};
void func(struct StructName struct_var){
printf("Param values are: %4X %4X\n", struct_var.s & 0xFFFF, struct_var.n & 0xFFFF);
}
int main(){
struct StructName struct_var;
struct_var.s = 0xDEAD;
struct_var.n = 0xBEEF;
func(struct_var);
}
//It looks like you are trying to use the definition as a variable. Here the definition is StructName and the variable is struct_var.
this sample code outputs:
Param values are: DEAD BEEF
If you use clang or gcc, you may be able to use typeof:
struct foo {
struct {
int i;
} anon;
} foo;
void do_something(typeof(foo.anon)* member) {
member->i = 1;
}
If there is no global instance of your type, you may be able to use typeof((struct foo){}.anon).
This comes with a lot of downsides. The most obvious ones are that:
it's not standard, and it ties you to clang/gcc
it's pretty darn ugly
it might not behave as you expect anyway
For instance, structurally-equivalent anonymous types do not have the same type, so in something like this:
struct foo {
struct {
int i;
} anon1;
struct {
int i;
} anon2;
} foo;
anon1 and anon2 both have a different type, meaning that typeof one of them cannot be used to refer to both.
In the long run, you will almost certainly find that it's worth naming the structures, especially if you use them as function arguments. For instance, if you want to make your variable available from a header, I think that you'll have to work pretty hard to keep it anonymous.
Although it's not particularly pretty and not compatible with C++, C puts the name of nested declarations in the global namespace, so this is portable and it's not a very big code change to front-load:
struct {
struct not_anon {
int i;
} anon;
} foo;
void do_something(struct not_anon* member) {
member->i = 1;
}
I have a structure where the first element is tested and dependent on its value the rest of the structure will or will not be read. In the cases where the first element's value dictates that the rest of the structure will not be read, do I have to allocate enough memory for the entire structure or just the first element?
struct element
{
int x;
int y;
};
int foo(struct element* e)
{
if(e->x > 3)
return e->y;
return e->x;
}
in main:
int i = 0;
int z = foo((struct element*)&i);
I assume that if only allocating for the first element is valid, then I will have to be wary of anything that may attempt to copy the structure. i.e. passing the struct to a function.
don't force your information into structs where it's not needed: don't use the struct as the parameter of your function.
either pass the member of your struct to the function or use inheritance:
typedef struct {
int foo;
} BaseA;
typedef struct {
int bar;
} BaseB;
typedef struct {
BaseA a;
BaseB b;
} Derived;
void foo(BaseB* info) { ... }
...
Derived d;
foo(&d.b);
BaseB b;
foo(&b);
if you're just curious (and seriously don't use this): you may.
typedef struct {
int foo, goo, hoo, joo;
} A;
typedef struct {
int unused, goo;
} B;
int foo(A* a) { return a->goo; }
...
B b;
int goo = foo((A*)&b);
In general you'll have to allocate a block of memory at least as many bytes as are required to fully read the accessed member with the largest offset in your structure. In addition when writing to this block you have to make sure to use the same member offsets as in the original structure.
The point being, a structure is only a block of memory with different areas assigned different interpretations (int, char, other structs etc...) and accessing a member of a struct (after reordering and alignment) boils down to simply reading from or writing to a bit of memory.
I do not think the code as given is legitimate. To understand why, consider:
struct CHAR_AND_INT { unsigned char c; int i; }
CHAR_AND_INT *p;
A compiler would be entitled to assume that p->c will be word-aligned and have whatever padding would be necessary for p->i to also be word-aligned. On some processors, writing a byte may be slower than writing a word. For example, a byte-store instruction may require the processor to read a word from memory, update one byte within it, and write the whole thing back, while a word-store instruction could simply store the new data without having to read anything first. A compiler that knew that p->c would be word-aligned and padded could implement p->c = 12; by using a word store to write the value 12. Such behavior wouldn't yield desired results, however, if the byte following p->c wasn't padding but instead held useful data.
While I would not expect a compiler to impose "special" alignment or padding requirements on any part of the structure shown in the original question (beyond those which apply to int) I don't think anything in the standard would forbid a compiler from doing so.
You need to only check that the structure itself is allocated; not the members (in that case at least)
int foo(struct element* e)
{
if ( e != 0) // check that the e pointer is valid
{
if(e->x != 0) // here you only check to see if x is different than zero (values, not pointers)
return e->y;
}
return 0;
}
In you edited change, I think this is poor coding
int i = 0;
int z = foo((struct element*)&i);
In that case, i will be allocation on the stack, so its address is valid; and will be valid in foo; but since you cast it into something different, the members will be garbage (at best)
Why do you want to cast an int into a structure?
What is your intent?
I'm trying to create structs with default values. I don't know how to accomplish this because every code that I see, is about initialising, and I would it for the natural way like...
struct stuff {
int stuff_a = 1;
int stuff_b = 2...
...and so on...
};
and looking about, I found this (C++) code:
struct a{ a() : i(0), j(0) {}; INT i; INT j;}
I never saw anything like this for C. Please, help me to understand it; I think that it is very nice!
UPDATE: Wait, I'm asking about C!!!! Why changed my question? If that is not possible in C just say... I don't know C++, I didn't know that was about C++...
If you want to set a struct object in one go and you have a C99 compiler, try this:
struct stuff {
int stuff_a;
int stuff_b;
// and so on...
};
struct stuff foo;
/* ... code ... */
foo = (struct stuff){.stuff_b = 42, .stuff_a = -1000};
Otherwise, with a C89 compiler, you have to set each member one by one:
foo.stuff_b = 42;
foo.stuff_a = -1000;
Running example # ideone : http://ideone.com/1QqCB
The original line
struct a{ a() : i(0), j(0) {} INT i; INT j;}
is a syntax error in C.
As you have probably learned from the other answers, in C you can't declare a structure and initialize it's members at the same time. These are different tasks and must be done separately.
There are a few options for initializing member variables of a struct. I'll show a couple of ways below. Right now, let's assume the following struct is defined in the beginning of the file:
struct stuff {
int stuff_a;
int stuff_b;
};
Then on your main() code, imagine that you want to declare a new variable of this type:
struct stuff custom_var;
This is the moment where you must initialize the structure. Seriously, I mean you really really must! Even if you don't want to assign specific values to them, you must at least initialize them to zero. This is mandatory because the OS doesn't guarantee that it will give you a clean memory space to run your application on. Therefore, always initialize your variables to some value (usually 0), including the other default types, such as char, int, float, double, etc...
One way to initialize our struct to zero is through memset():
memset(&custom_var, 0, sizeof(struct stuff));
Another is accessing each member individually:
custom_var.stuff_a = 0;
custom_var.stuff_b = 0;
A third option, which might confuse beginners is when they see the initialization of struct members being done at the moment of the declaration:
struct stuff custom_var = { 1, 2 };
The code above is equivalent to:
struct stuff custom_var;
custom_var.stuff_a = 1;
custom_var.stuff_b = 2;
... create structs with default values ...
That is impossible in C. A type cannot have default values. Objects of any type cannot have a default value other than 0, though they can be initialized to whatever is wanted.
The definition of a struct is a definition of a type, not of an object.
What you asking is about the same thing as a way to have ints default to, say, 42.
/* WRONG CODE -- THIS DOES NOT WORK */
typedef int int42 = 42;
int42 a;
printf("%d\n", a); /* print 42 */
Or, adapting to your example
/* WRONG CODE -- THIS DOES NOT WORK */
struct stuff {
int42 stuff_a;
int65536 stuff_b;
}
struct stuff a;
printf("%d\n", a.stuff_b); /* print 65536 */
Update: This answer assumes we 're talking about C++ because the code posted in the answer is not legal C.
struct a {
a() : i(0), j(0) {} // constructor with initialization list
int i;
int j;
}
The line marked with the comment is simply the constructor for instances of struct a (reminder: structs are just like classes, except that the default member visibility is public instead of private).
The part after the : is called an initialization list: it allows you to initialize the members of the struct with values (either constants or passed as constructor parameters). Initialization of members in this list happens before the body of the constructor is entered. It is preferable to initialize members of classes and structs this way, if at all possible.
See also C++: Constructor versus initializer list in struct/class.
in C (pre C99) the following also works:
#include <stdio.h>
typedef struct
{
int a;
int b;
int c;
} HELLO;
int main()
{
HELLO a = {1,2,3};
printf("here: %d %d %d\n",a.a,a.b,a.c);
exit(1);
}
See codepad
I'm not sure quite sure what your problem is. The standard way of initialising structures in c is like this:
struct a_struct my_struct = {1, 2};
Or the more recent and safer:
struct a_struct my_struct = {.i1 = 1, .i2 = 2};
If there is more than one instance of a structure, or it needs to be re-initialised, it is useful to define a constant structure with default values then assign that.
typedef struct a_struct {
int i1;
int i2;
} sa;
static const sa default_sa = {.i1 = 1, .i2 = 2};
static sa sa1 = default_sa;
static sa sa2 = default_sa;
// obviously you can do it dynamically as well
void use_temp_sa(void)
{
sa temp_sa = default_sa;
temp_sa.i2 = 3;
do_something_with(&temp_sa);
}
// And re-initialise
void reset_sa(sa *my_sa)
{
*my_sa = default_sa;
}
Type initializer is not possible in C.
A value must be stored in the memory.
A type does not occupy memory, what occupies memory is a variable of that type.
struct stuff; is a type; it does not occupy memory
struct stuff aStuff; is a variable of that type; aStuff occupies memory
Because a type does not occupy memory, it is not possible to save values into a type.
If there is syntactic sugar to support store/initialize values into a type then there must be additional code that is inserted to assign values to every instant variables of that type (e.g: in constructor in C++). This will result in a less efficient C if this feature is available.
How often do you need to retain this default values? I think it is unlikely. You can create a function to initialize variable with the default values or just initialize every fields with the values you want. So type initializer is not fundamental thing. C is about simplicity.
Can't initialize values within a structure definition.
I'd suggest:
typedef struct {
int stuff_a;
int stuff_b;
} stuff ;
int stuffInit(int a, int b, stuff *this){
this->stuff_a = a;
this->stuff_b = b;
return 0; /*or an error code, or sometimes '*this', per taste.*/
}
int main(void){
stuff myStuff;
stuffInit(1, 2, &myStuff);
/* dynamic is more commonly seen */
stuff *dynamicStuff;
dynamicStuff = malloc(sizeof(stuff)); /* 'new' stuff */
stuffInit(0, 0, dynamicStuff);
free(dynamicStuff); /* 'delete' stuff */
return 0;
}
Before the days of Object Oriented Programming (C++), we were taught "Abstract Data Types".
The discipline said 'never access your data structures directly, always create a function for it' But this was only enforced by the programmer, instructor, or senior developer, not the language.
Eventually, the structure definition(s) and corresponding functions end up in their own file & header, linked in later, further encapsulating the design.
But those days are gone and replaced with 'Class' and 'Constructor' OOP terminology.
"It's all the same, only the names have changed" - Bon Jovi.
I have some code with multiple functions very similar to each other to look up an item in a list based on the contents of one field in a structure. The only difference between the functions is the type of the structure that the look up is occurring in. If I could pass in the type, I could remove all the code duplication.
I also noticed that there is some mutex locking happening in these functions as well, so I think I might leave them alone...
If you ensure that the field is placed in the same place in each such structure, you can simply cast a pointer to get at the field. This technique is used in lots of low level system libraries e.g. BSD sockets.
struct person {
int index;
};
struct clown {
int index;
char *hat;
};
/* we're not going to define a firetruck here */
struct firetruck;
struct fireman {
int index;
struct firetruck *truck;
};
int getindexof(struct person *who)
{
return who->index;
}
int main(int argc, char *argv[])
{
struct fireman sam;
/* somehow sam gets initialised */
sam.index = 5;
int index = getindexof((struct person *) &sam);
printf("Sam's index is %d\n", index);
return 0;
}
You lose type safety by doing this, but it's a valuable technique.
[ I have now actually tested the above code and fixed the various minor errors. It's much easier when you have a compiler. ]
Since structures are nothing more than predefined blocks of memory, you can do this. You could pass a void * to the structure, and an integer or something to define the type.
From there, the safest thing to do would be to recast the void * into a pointer of the appropriate type before accessing the data.
You'll need to be very, very careful, as you lose type-safety when you cast to a void * and you can likely end up with a difficult to debug runtime error when doing something like this.
I think you should look at the C standard functions qsort() and bsearch() for inspiration. These are general purpose code to sort arrays and to search for data in a pre-sorted array. They work on any type of data structure - but you pass them a pointer to a helper function that does the comparisons. The helper function knows the details of the structure, and therefore does the comparison correctly.
In fact, since you are wanting to do searches, it may be that all you need is bsearch(), though if you are building the data structures on the fly, you may decide you need a different structure than a sorted list. (You can use sorted lists -- it just tends to slow things down compared with, say, a heap. However, you'd need a general heap_search() function, and a heap_insert() function, to do the job properly, and such functions are not standardized in C. Searching the web shows such functions exist - not by that name; just do not try "c heap search" since it is assumed you meant "cheap search" and you get tons of junk!)
If the ID field you test is part of a common initial sequence of fields shared by all the structs, then using a union guarantees that the access will work:
#include <stdio.h>
typedef struct
{
int id;
int junk1;
} Foo;
typedef struct
{
int id;
long junk2;
} Bar;
typedef union
{
struct
{
int id;
} common;
Foo foo;
Bar bar;
} U;
int matches(const U *candidate, int wanted)
{
return candidate->common.id == wanted;
}
int main(void)
{
Foo f = { 23, 0 };
Bar b = { 42, 0 };
U fu;
U bu;
fu.foo = f;
bu.bar = b;
puts(matches(&fu, 23) ? "true" : "false");
puts(matches(&bu, 42) ? "true" : "false");
return 0;
}
If you're unlucky, and the field appears at different offsets in the various structs, you can add an offset parameter to your function. Then, offsetof and a wrapper macro simulate what the OP asked for - passing the type of struct at the call site:
#include <stddef.h>
#include <stdio.h>
typedef struct
{
int id;
int junk1;
} Foo;
typedef struct
{
int junk2;
int id;
} Bar;
int matches(const void* candidate, size_t idOffset, int wanted)
{
return *(int*)((const unsigned char*)candidate + idOffset) == wanted;
}
#define MATCHES(type, candidate, wanted) matches(candidate, offsetof(type, id), wanted)
int main(void)
{
Foo f = { 23, 0 };
Bar b = { 0, 42 };
puts(MATCHES(Foo, &f, 23) ? "true" : "false");
puts(MATCHES(Bar, &b, 42) ? "true" : "false");
return 0;
}
One way to do this is to have a type field as the first byte of the structure. Your receiving function looks at this byte and then casts the pointer to the correct type based on what it discovers. Another approach is to pass the type information as a separate parameter to each function that needs it.
You can do this with a parameterized macro but most coding policies will frown on that.
#include
#define getfield(s, name) ((s).name)
typedef struct{
int x;
}Bob;
typedef struct{
int y;
}Fred;
int main(int argc, char**argv){
Bob b;
b.x=6;
Fred f;
f.y=7;
printf("%d, %d\n", getfield(b, x), getfield(f, y));
}
Short answer: no. You can, however, create your own method for doing so, i.e. providing a specification for how to create such a struct. However, it's generally not necessary and is not worth the effort; just pass by reference. (callFuncWithInputThenOutput(input, &struct.output);)
I'm a little rusty on c, but try using a void* pointer as the variable type in the function parameter. Then pass the address of the structure to the function, and then use it he way that you would.
void foo(void* obj);
void main()
{
struct bla obj;
...
foo(&obj);
...
}
void foo(void* obj)
{
printf(obj -> x, "%s")
}