I want to pass an array of arbitrary struct pointers and a comparison function to a generic sorting algorithm. Is this possible in C?
The goooeys of the structs would only be accessed within the comparison function, the sorting function would only need to call the comparison function and swap pointers, but I can't figure out how to declare it.
function sorter( struct arbitrary ** Array, int Length, int cmp(struct node * a, struct node * b))
{
for (int i=0; i<Length;i++){
if cmp(Array[i],Array[i+1]){
swap(Array[i],Array[i+1]
}
}
}
You can declare the function as:
void sorter(void** the_array, size_t array_length, int (*comparison_function)(void*, void*));
Inside of the comparison function, you will then need to cast the two pointers being compared to pointers to whatever struct type the comparison function compares.
Actually, this function already exists... it is called qsort. See some documentation here. It's also more efficient than your implementation, which is O(n^2).
Maybe you need to pass just void pointers?
function sorter(void ** Array, int Length, int cmp(void * a, void * b))
It's always possible in C simply because you can convert every pointer to a void*. But, if you want to be able to convert that back to a pointer to an arbitrary structure, you'll need some sort of type identification.
You could do that by using a function specific to the type (if the things you are comparing are the same), or encode the type into the structure somehow. This can be done by having an extra field in the structure or by changing the cmp() function itself to take a type identifier.
But you should be aware that C already has a qsort() function that's usually reasonably efficient (although there's nothing in the standard that dictates what algorithm it uses - it could use bubble sort and still be conforming). Unless you're implementing one for homework, or have a different algorithm in mind, tou should probably just use that.
Your algorithm, as it stands, looks like the inner loop of a bubble sort and, as such, won't actually sort correctly. Bubble sort consists of two nested loops and is generally only suitable for small data sets or those with specific characteristics (such as mostly sorted already).
Related
This question goes out to the C gurus out there:
In C, it is possible to declare a pointer as follows:
char (* p)[10];
.. which basically states that this pointer points to an array of 10 chars. The neat thing about declaring a pointer like this is that you will get a compile time error if you try to assign a pointer of an array of different size to p. It will also give you a compile time error if you try to assign the value of a simple char pointer to p. I tried this with gcc and it seems to work with ANSI, C89 and C99.
It looks to me like declaring a pointer like this would be very useful - particularly, when passing a pointer to a function. Usually, people would write the prototype of such a function like this:
void foo(char * p, int plen);
If you were expecting a buffer of an specific size, you would simply test the value of plen. However, you cannot be guaranteed that the person who passes p to you will really give you plen valid memory locations in that buffer. You have to trust that the person who called this function is doing the right thing. On the other hand:
void foo(char (*p)[10]);
..would force the caller to give you a buffer of the specified size.
This seems very useful but I have never seen a pointer declared like this in any code I have ever ran across.
My question is: Is there any reason why people do not declare pointers like this? Am I not seeing some obvious pitfall?
What you are saying in your post is absolutely correct. I'd say that every C developer comes to exactly the same discovery and to exactly the same conclusion when (if) they reach certain level of proficiency with C language.
When the specifics of your application area call for an array of specific fixed size (array size is a compile-time constant), the only proper way to pass such an array to a function is by using a pointer-to-array parameter
void foo(char (*p)[10]);
(in C++ language this is also done with references
void foo(char (&p)[10]);
).
This will enable language-level type checking, which will make sure that the array of exactly correct size is supplied as an argument. In fact, in many cases people use this technique implicitly, without even realizing it, hiding the array type behind a typedef name
typedef int Vector3d[3];
void transform(Vector3d *vector);
/* equivalent to `void transform(int (*vector)[3])` */
...
Vector3d vec;
...
transform(&vec);
Note additionally that the above code is invariant with relation to Vector3d type being an array or a struct. You can switch the definition of Vector3d at any time from an array to a struct and back, and you won't have to change the function declaration. In either case the functions will receive an aggregate object "by reference" (there are exceptions to this, but within the context of this discussion this is true).
However, you won't see this method of array passing used explicitly too often, simply because too many people get confused by a rather convoluted syntax and are simply not comfortable enough with such features of C language to use them properly. For this reason, in average real life, passing an array as a pointer to its first element is a more popular approach. It just looks "simpler".
But in reality, using the pointer to the first element for array passing is a very niche technique, a trick, which serves a very specific purpose: its one and only purpose is to facilitate passing arrays of different size (i.e. run-time size). If you really need to be able to process arrays of run-time size, then the proper way to pass such an array is by a pointer to its first element with the concrete size supplied by an additional parameter
void foo(char p[], unsigned plen);
Actually, in many cases it is very useful to be able to process arrays of run-time size, which also contributes to the popularity of the method. Many C developers simply never encounter (or never recognize) the need to process a fixed-size array, thus remaining oblivious to the proper fixed-size technique.
Nevertheless, if the array size is fixed, passing it as a pointer to an element
void foo(char p[])
is a major technique-level error, which unfortunately is rather widespread these days. A pointer-to-array technique is a much better approach in such cases.
Another reason that might hinder the adoption of the fixed-size array passing technique is the dominance of naive approach to typing of dynamically allocated arrays. For example, if the program calls for fixed arrays of type char[10] (as in your example), an average developer will malloc such arrays as
char *p = malloc(10 * sizeof *p);
This array cannot be passed to a function declared as
void foo(char (*p)[10]);
which confuses the average developer and makes them abandon the fixed-size parameter declaration without giving it a further thought. In reality though, the root of the problem lies in the naive malloc approach. The malloc format shown above should be reserved for arrays of run-time size. If the array type has compile-time size, a better way to malloc it would look as follows
char (*p)[10] = malloc(sizeof *p);
This, of course, can be easily passed to the above declared foo
foo(p);
and the compiler will perform the proper type checking. But again, this is overly confusing to an unprepared C developer, which is why you won't see it in too often in the "typical" average everyday code.
I would like to add to AndreyT's answer (in case anyone stumbles upon this page looking for more info on this topic):
As I begin to play more with these declarations, I realize that there is major handicap associated with them in C (apparently not in C++). It is fairly common to have a situation where you would like to give a caller a const pointer to a buffer you have written into. Unfortunately, this is not possible when declaring a pointer like this in C. In other words, the C standard (6.7.3 - Paragraph 8) is at odds with something like this:
int array[9];
const int (* p2)[9] = &array; /* Not legal unless array is const as well */
This constraint does not seem to be present in C++, making these type of declarations far more useful. But in the case of C, it is necessary to fall back to a regular pointer declaration whenever you want a const pointer to the fixed size buffer (unless the buffer itself was declared const to begin with). You can find more info in this mail thread: link text
This is a severe constraint in my opinion and it could be one of the main reasons why people do not usually declare pointers like this in C. The other being the fact that most people do not even know that you can declare a pointer like this as AndreyT has pointed out.
The obvious reason is that this code doesn't compile:
extern void foo(char (*p)[10]);
void bar() {
char p[10];
foo(p);
}
The default promotion of an array is to an unqualified pointer.
Also see this question, using foo(&p) should work.
I also want to use this syntax to enable more type checking.
But I also agree that the syntax and mental model of using pointers is simpler, and easier to remember.
Here are some more obstacles I have come across.
Accessing the array requires using (*p)[]:
void foo(char (*p)[10])
{
char c = (*p)[3];
(*p)[0] = 1;
}
It is tempting to use a local pointer-to-char instead:
void foo(char (*p)[10])
{
char *cp = (char *)p;
char c = cp[3];
cp[0] = 1;
}
But this would partially defeat the purpose of using the correct type.
One has to remember to use the address-of operator when assigning an array's address to a pointer-to-array:
char a[10];
char (*p)[10] = &a;
The address-of operator gets the address of the whole array in &a, with the correct type to assign it to p. Without the operator, a is automatically converted to the address of the first element of the array, same as in &a[0], which has a different type.
Since this automatic conversion is already taking place, I am always puzzled that the & is necessary. It is consistent with the use of & on variables of other types, but I have to remember that an array is special and that I need the & to get the correct type of address, even though the address value is the same.
One reason for my problem may be that I learned K&R C back in the 80s, which did not allow using the & operator on whole arrays yet (although some compilers ignored that or tolerated the syntax). Which, by the way, may be another reason why pointers-to-arrays have a hard time to get adopted: they only work properly since ANSI C, and the & operator limitation may have been another reason to deem them too awkward.
When typedef is not used to create a type for the pointer-to-array (in a common header file), then a global pointer-to-array needs a more complicated extern declaration to share it across files:
fileA:
char (*p)[10];
fileB:
extern char (*p)[10];
Well, simply put, C doesn't do things that way. An array of type T is passed around as a pointer to the first T in the array, and that's all you get.
This allows for some cool and elegant algorithms, such as looping through the array with expressions like
*dst++ = *src++
The downside is that management of the size is up to you. Unfortunately, failure to do this conscientiously has also led to millions of bugs in C coding, and/or opportunities for malevolent exploitation.
What comes close to what you ask in C is to pass around a struct (by value) or a pointer to one (by reference). As long as the same struct type is used on both sides of this operation, both the code that hand out the reference and the code that uses it are in agreement about the size of the data being handled.
Your struct can contain whatever data you want; it could contain your array of a well-defined size.
Still, nothing prevents you or an incompetent or malevolent coder from using casts to fool the compiler into treating your struct as one of a different size. The almost unshackled ability to do this kind of thing is a part of C's design.
You can declare an array of characters a number of ways:
char p[10];
char* p = (char*)malloc(10 * sizeof(char));
The prototype to a function that takes an array by value is:
void foo(char* p); //cannot modify p
or by reference:
void foo(char** p); //can modify p, derefernce by *p[0] = 'f';
or by array syntax:
void foo(char p[]); //same as char*
I would not recommend this solution
typedef int Vector3d[3];
since it obscures the fact that Vector3D has a type that you
must know about. Programmers usually dont expect variables of the
same type to have different sizes. Consider :
void foo(Vector3d a) {
Vector3d b;
}
where sizeof a != sizeof b
Maybe I'm missing something, but... since arrays are constant pointers, basically that means that there's no point in passing around pointers to them.
Couldn't you just use void foo(char p[10], int plen); ?
type (*)[];
// points to an array e.g
int (*ptr)[5];
// points to an 5 integer array
// gets the address of the array
type *[];
// points to an array of pointers e.g
int* ptr[5]
// point to an array of five integer pointers
// point to 5 adresses.
On my compiler (vs2008) it treats char (*p)[10] as an array of character pointers, as if there was no parentheses, even if I compile as a C file. Is compiler support for this "variable"? If so that is a major reason not to use it.
before you mark this as a duplicate please notice that I'm looking for a more general solution for arrays of arbitrary dimensions. I have read many posts here or in forums about making 2D or 3D arrays of integers but these are specific solutions for specific dimensions. I want a general solution for an array of any dimension.
First I need to have a type of intlist as defined below:
typedef struct{
int l // length of the list
int * e // pointer to the first element of the array
}intlist;
this actually fills the gap in C for treating arrays just as pointers. using this type I can pass arrays to functions without worrying about loosing the size.
then in the next step I want to have a mdintlist as multidimensional dynamically allocated arrays. the type definition should be something like this:
typedef struct Mdintlist{
intlist d // dimension of the array
/* second part */
}mdintlist;
there are several options for the second part. on option is that to have a pointer towards a mdintlist of lower dimension like
struct Mdintlist * c;
the other options is to use void pointers:
void * c;
I don't know how to continue it from here.
P.S. one solution could be to allocate just one block of memory and then call the elements using a function. However I would like to call the elements in array form. something like tmpmdintlist.c[1][2][3]...
Hope I have explained clearly what I want.
P.S. This is an ancient post, but for those who may end up here some of my efforts can be seen in the Cplus repo.
You can't! you can only use the function option in c, because there is no way to alter the language semantics. In c++ however you can overload the [] operator, and even though I would never do such an ugly thing (x[1][2][3] is alread y ugly, if you continue adding "dimensions" it gets really ugly), I think it would be possible.
Well, if you separate the pointers and the array lengths, you end up with much less code.
int *one_dem_array;
size_t one_dem_count[1];
int **two_dem_array;
size_t two_dem_count[2];
int ***three_dem_array;
size_t three_dem_count[3];
This way you can still use your preferred notation.
int num_at_pos = three_dem_array[4][2][3];
C newbie here, I need some help: Can anyone explain to (and offer a workaroud) me why this works:
int n=1024;
int32_t data[n];
void synthesize_signal(int32_t *data) {
...//do something with data}
which let me alter data in the function; but this does not?
int n=1024;
int number=1024*16;
int32_t data[n][2][number];
void synthesize_signal(int32_t *data) {
...//do something with data}
The compiler error message is something like it expected int32_t * but got int32_t (*)[2][(sizetype)(number)] instead.
First, passing arrays in C is by reference. So you pass a pointer of some sort, and the function can modify the data in the array. You don't have to worry about passing a pointer to the array. In fact, in C there is no real different between a pointer that happens to be to the being of an array, and the array itself.
In your first version. You making a one-dimensional array data[n], and you are passing it to your function. In the array, you'll using it by saying, something like data[i]. This translates directly to (data + (i sizeof(int32_t)). It is using the size of the elements in the array to find the memory location that is i positions in front of the beginning of your array.
int n=1024;
int number=1024*16;
int32_t data[n][2][number];
void synthesize_signal(int32_t *data)
In the second case, you're setting up a mufti-dimensional array (3D in your case). You setup correctly. The problem is that when you pass it to the function, the only thing that gets passed the address of the being of the array. When it gets used inside the function, you'll do something like
data[i][1][x] = 5;
Internally C is calculating how from the beginning of the array this location is. In order for it to do that, it need to know the dimensions of the array. (Unlike some newer languages, C store any extra data about array lengths or sizes or anything). You just need to change the function signature so it knows the shape/size of array to expect. Because of the way, it calculates array positions, it doesn't need the first dimension.
In this case, change your function signature to look like this:
void synthesize_signal(int32_t data[][2][number]) { ...
Setup the array the same way you are doing the second one above, and just call it you'd expect:
synthesize_signal(data);
This should fix everything for you.
The comments mention some useful information about using more descriptive variable names, and global vs. local variable. All valid comments to keep in mind. I just addressed to code problem you're having in terms of mufti-dimensional arrays.
try
synthesize_signal(int32_t** data)
{
}
Your function also needs to know that data is multi dimensional. You should also consider renaming your data array. I suspect that it is a global variable and using the same name in function can lead to problems.
When you call the function, do it like this:
synthesize_signal(&data[0][0][0]);
I feel like triple pointers in C are looked at as "bad". For me, it makes sense to use them at times.
Starting from the basics, the single pointer has two purposes: to create an array, and to allow a function to change its contents (pass by reference):
char *a;
a = malloc...
or
void foo (char *c); //means I'm going to modify the parameter in foo.
{ *c = 'f'; }
char a;
foo(&a);
The double pointer can be a 2D array (or array of arrays, since each "column" or "row" need not be the same length). I personally like to use it when I need to pass a 1D array:
void foo (char **c); //means I'm going to modify the elements of an array in foo.
{ (*c)[0] = 'f'; }
char *a;
a = malloc...
foo(&a);
To me, that helps describe what foo is doing. However, it is not necessary:
void foo (char *c); //am I modifying a char or just passing a char array?
{ c[0] = 'f'; }
char *a;
a = malloc...
foo(a);
will also work.
According to the first answer to this question, if foo were to modify the size of the array, a double pointer would be required.
One can clearly see how a triple pointer (and beyond, really) would be required. In my case if I were passing an array of pointers (or array of arrays), I would use it. Evidently it would be required if you are passing into a function that is changing the size of the multi-dimensional array. Certainly an array of arrays of arrays is not too common, but the other cases are.
So what are some of the conventions out there? Is this really just a question of style/readability combined with the fact that many people have a hard time wrapping their heads around pointers?
Using triple+ pointers is harming both readability and maintainability.
Let's suppose you have a little function declaration here:
void fun(int***);
Hmmm. Is the argument a three-dimensional jagged array, or pointer to two-dimensional jagged array, or pointer to pointer to array (as in, function allocates an array and assigns a pointer to int within a function)
Let's compare this to:
void fun(IntMatrix*);
Surely you can use triple pointers to int to operate on matrices. But that's not what they are. The fact that they're implemented here as triple pointers is irrelevant to the user.
Complicated data structures should be encapsulated. This is one of manifest ideas of Object Oriented Programming. Even in C, you can apply this principle to some extent. Wrap the data structure in a struct (or, very common in C, using "handles", that is, pointers to incomplete type - this idiom will be explained later in the answer).
Let's suppose that you implemented the matrices as jagged arrays of double. Compared to contiguous 2D arrays, they are worse when iterating over them (as they don't belong to a single block of contiguous memory) but allow for accessing with array notation and each row can have different size.
So now the problem is you can't change representations now, as the usage of pointers is hard-wired over user code, and now you're stuck with inferior implementation.
This wouldn't be even a problem if you encapsulated it in a struct.
typedef struct Matrix_
{
double** data;
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i][j];
}
simply gets changed to
typedef struct Matrix_
{
int width;
double data[]; //C99 flexible array member
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i*m->width+j];
}
The handle technique works like this: in the header file, you declare a incomplete struct and all the functions that work on the pointer to the struct:
// struct declaration with no body.
struct Matrix_;
// optional: allow people to declare the matrix with Matrix* instead of struct Matrix*
typedef struct Matrix_ Matrix;
Matrix* create_matrix(int w, int h);
void destroy_matrix(Matrix* m);
double get_element(Matrix* m, int i, int j);
double set_element(Matrix* m, double value, int i, int j);
in the source file you declare the actual struct and define all the functions:
typedef struct Matrix_
{
int width;
double data[]; //C99 flexible array member
} Matrix;
double get_element(Matrix* m, int i, int j)
{
return m->data[i*m->width+j];
}
/* definition of the rest of the functions */
The rest of the world doesn't know what does the struct Matrix_ contain and it doesn't know the size of it. This means users can't declare the values directly, but only by using pointer to Matrix and the create_matrix function. However, the fact that the user doesn't know the size means the user doesn't depend on it - which means we can remove or add members to struct Matrix_ at will.
Most of the time, the use of 3 levels of indirection is a symptom of bad design decisions made elsewhere in the program. Therefore it is regarded as bad practice and there are jokes about "three star programmers" where, unlike the the rating for restaurants, more stars means worse quality.
The need for 3 levels of indirection often originates from the confusion about how to properly allocate multi-dimensional arrays dynamically. This is often taught incorrectly even in programming books, partially because doing it correctly was burdensome before the C99 standard. My Q&A post Correctly allocating multi-dimensional arrays addresses that very issue and also illustrates how multiple levels of indirection will make the code increasingly hard to read and maintain.
Though as that post explains, there are some situations where a type** might make sense. A variable table of strings with variable length is such an example. And when that need for type** arises, you might soon be tempted to use type***, because you need to return your type** through a function parameter.
Most often this need arises in a situation where you are designing some manner of complex ADT. For example, lets say that we are coding a hash table, where each index is a 'chained' linked list, and each node in the linked list an array. The proper solution then is to re-design the program to use structs instead of multiple levels of indirection. The hash table, linked list and array should be distinct types, autonomous types without any awareness of each other.
So by using proper design, we will avoid the multiple stars automatically.
But as with every rule of good programming practice, there are always exceptions. It is perfectly possible to have a situation like:
Must implement an array of strings.
The number of strings is variable and may change in run-time.
The length of the strings is variable.
You can implement the above as an ADT, but there may also be valid reasons to keep things simple and just use a char* [n]. You then have two options to allocate this dynamically:
char* (*arr_ptr)[n] = malloc( sizeof(char*[n]) );
or
char** ptr_ptr = malloc( sizeof(char*[n]) );
The former is more formally correct, but also cumbersome. Because it has to be used as (*arr_ptr)[i] = "string";, while the alternative can be used as ptr_ptr[i] = "string";.
Now suppose we have to place the malloc call inside a function and the return type is reserved for an error code, as is custom with C APIs. The two alternatives will then look like this:
err_t alloc_arr_ptr (size_t n, char* (**arr)[n])
{
*arr = malloc( sizeof(char*[n]) );
return *arr == NULL ? ERR_ALLOC : OK;
}
or
err_t alloc_ptr_ptr (size_t n, char*** arr)
{
*arr = malloc( sizeof(char*[n]) );
return *arr == NULL ? ERR_ALLOC : OK;
}
It is quite hard to argue and say that the former is more readable, and it also comes with the cumbersome access needed by the caller. The three star alternative is actually more elegant, in this very specific case.
So it does us no good to dismiss 3 levels of indirection dogmatically. But the choice to use them must be well-informed, with an awareness that they may create ugly code and that there are other alternatives.
So what are some of the conventions out there? Is this really just a question of style/readability combined with the fact that many people have a hard time wrapping their heads around pointers?
Multiple indirection is not bad style, nor black magic, and if you're dealing with high-dimension data then you're going to be dealing with high levels of indirection; if you're really dealing with a pointer to a pointer to a pointer to T, then don't be afraid to write T ***p;. Don't hide pointers behind typedefs unless whoever is using the type doesn't have to worry about its "pointer-ness". For example, if you're providing the type as a "handle" that gets passed around in an API, such as:
typedef ... *Handle;
Handle h = NewHandle();
DoSomethingWith( h, some_data );
DoSomethingElseWith( h, more_data );
ReleaseHandle( h );
then sure, typedef away. But if h is ever meant to be dereferenced, such as
printf( "Handle value is %d\n", *h );
then don't typedef it. If your user has to know that h is a pointer to int1 in order to use it properly, then that information should not be hidden behind a typedef.
I will say that in my experience I haven't had to deal with higher levels of indirection; triple indirection has been the highest, and I haven't had to use it more than a couple of times. If you regularly find yourself dealing with >3-dimensional data, then you'll see high levels of indirection, but if you understand how pointer expressions and indirection work it shouldn't be an issue.
1. Or a pointer to pointer to int, or pointer to pointer to pointer to pointer to struct grdlphmp, or whatever.
After two levels of indirection, comprehension becomes difficult. Moreover if the reason you're passing these triple (or more) pointers into your methods is so that they can re-allocate and re-set some pointed-to memory, that gets away from the concept of methods as "functions" that just return values and don't affect state. This also negatively affects comprehension and maintainability beyond some point.
But more fundamentally, you've hit upon one of the main stylistic objections to the triple pointer right here:
One can clearly see how a triple pointer (and beyond, really) would be required.
It's the "and beyond" that is the issue here: once you get to three levels, where do you stop? Surely it's possible to have an aribitrary number of levels of indirection. But it's better to just have a customary limit someplace where comprehensibility is still good but flexibility is adequate. Two's a good number. "Three star programming", as it's sometimes called, is controversial at best; it's either brilliant, or a headache for those who need to maintain the code later.
Unfortunately you misunderstood the concept of pointer and arrays in C. Remember that arrays are not pointers.
Starting from the basics, the single pointer has two purposes: to create an array, and to allow a function to change its contents (pass by reference):
When you declare a pointer, then you need to initialize it before using it in the program. It can be done either by passing address of a variable to it or by dynamic memory allocation.
In latter, pointer can be used as indexed arrays (but it is not an array).
The double pointer can be a 2D array (or array of arrays, since each "column" or "row" need not be the same length). I personally like to use it when I need to pass a 1D array:
Again wrong. Arrays are not pointers and vice-versa. A pointer to pointer is not the 2D array.
I would suggest you to read the c-faq section 6. Arrays and Pointers.
This question goes out to the C gurus out there:
In C, it is possible to declare a pointer as follows:
char (* p)[10];
.. which basically states that this pointer points to an array of 10 chars. The neat thing about declaring a pointer like this is that you will get a compile time error if you try to assign a pointer of an array of different size to p. It will also give you a compile time error if you try to assign the value of a simple char pointer to p. I tried this with gcc and it seems to work with ANSI, C89 and C99.
It looks to me like declaring a pointer like this would be very useful - particularly, when passing a pointer to a function. Usually, people would write the prototype of such a function like this:
void foo(char * p, int plen);
If you were expecting a buffer of an specific size, you would simply test the value of plen. However, you cannot be guaranteed that the person who passes p to you will really give you plen valid memory locations in that buffer. You have to trust that the person who called this function is doing the right thing. On the other hand:
void foo(char (*p)[10]);
..would force the caller to give you a buffer of the specified size.
This seems very useful but I have never seen a pointer declared like this in any code I have ever ran across.
My question is: Is there any reason why people do not declare pointers like this? Am I not seeing some obvious pitfall?
What you are saying in your post is absolutely correct. I'd say that every C developer comes to exactly the same discovery and to exactly the same conclusion when (if) they reach certain level of proficiency with C language.
When the specifics of your application area call for an array of specific fixed size (array size is a compile-time constant), the only proper way to pass such an array to a function is by using a pointer-to-array parameter
void foo(char (*p)[10]);
(in C++ language this is also done with references
void foo(char (&p)[10]);
).
This will enable language-level type checking, which will make sure that the array of exactly correct size is supplied as an argument. In fact, in many cases people use this technique implicitly, without even realizing it, hiding the array type behind a typedef name
typedef int Vector3d[3];
void transform(Vector3d *vector);
/* equivalent to `void transform(int (*vector)[3])` */
...
Vector3d vec;
...
transform(&vec);
Note additionally that the above code is invariant with relation to Vector3d type being an array or a struct. You can switch the definition of Vector3d at any time from an array to a struct and back, and you won't have to change the function declaration. In either case the functions will receive an aggregate object "by reference" (there are exceptions to this, but within the context of this discussion this is true).
However, you won't see this method of array passing used explicitly too often, simply because too many people get confused by a rather convoluted syntax and are simply not comfortable enough with such features of C language to use them properly. For this reason, in average real life, passing an array as a pointer to its first element is a more popular approach. It just looks "simpler".
But in reality, using the pointer to the first element for array passing is a very niche technique, a trick, which serves a very specific purpose: its one and only purpose is to facilitate passing arrays of different size (i.e. run-time size). If you really need to be able to process arrays of run-time size, then the proper way to pass such an array is by a pointer to its first element with the concrete size supplied by an additional parameter
void foo(char p[], unsigned plen);
Actually, in many cases it is very useful to be able to process arrays of run-time size, which also contributes to the popularity of the method. Many C developers simply never encounter (or never recognize) the need to process a fixed-size array, thus remaining oblivious to the proper fixed-size technique.
Nevertheless, if the array size is fixed, passing it as a pointer to an element
void foo(char p[])
is a major technique-level error, which unfortunately is rather widespread these days. A pointer-to-array technique is a much better approach in such cases.
Another reason that might hinder the adoption of the fixed-size array passing technique is the dominance of naive approach to typing of dynamically allocated arrays. For example, if the program calls for fixed arrays of type char[10] (as in your example), an average developer will malloc such arrays as
char *p = malloc(10 * sizeof *p);
This array cannot be passed to a function declared as
void foo(char (*p)[10]);
which confuses the average developer and makes them abandon the fixed-size parameter declaration without giving it a further thought. In reality though, the root of the problem lies in the naive malloc approach. The malloc format shown above should be reserved for arrays of run-time size. If the array type has compile-time size, a better way to malloc it would look as follows
char (*p)[10] = malloc(sizeof *p);
This, of course, can be easily passed to the above declared foo
foo(p);
and the compiler will perform the proper type checking. But again, this is overly confusing to an unprepared C developer, which is why you won't see it in too often in the "typical" average everyday code.
I would like to add to AndreyT's answer (in case anyone stumbles upon this page looking for more info on this topic):
As I begin to play more with these declarations, I realize that there is major handicap associated with them in C (apparently not in C++). It is fairly common to have a situation where you would like to give a caller a const pointer to a buffer you have written into. Unfortunately, this is not possible when declaring a pointer like this in C. In other words, the C standard (6.7.3 - Paragraph 8) is at odds with something like this:
int array[9];
const int (* p2)[9] = &array; /* Not legal unless array is const as well */
This constraint does not seem to be present in C++, making these type of declarations far more useful. But in the case of C, it is necessary to fall back to a regular pointer declaration whenever you want a const pointer to the fixed size buffer (unless the buffer itself was declared const to begin with). You can find more info in this mail thread: link text
This is a severe constraint in my opinion and it could be one of the main reasons why people do not usually declare pointers like this in C. The other being the fact that most people do not even know that you can declare a pointer like this as AndreyT has pointed out.
The obvious reason is that this code doesn't compile:
extern void foo(char (*p)[10]);
void bar() {
char p[10];
foo(p);
}
The default promotion of an array is to an unqualified pointer.
Also see this question, using foo(&p) should work.
I also want to use this syntax to enable more type checking.
But I also agree that the syntax and mental model of using pointers is simpler, and easier to remember.
Here are some more obstacles I have come across.
Accessing the array requires using (*p)[]:
void foo(char (*p)[10])
{
char c = (*p)[3];
(*p)[0] = 1;
}
It is tempting to use a local pointer-to-char instead:
void foo(char (*p)[10])
{
char *cp = (char *)p;
char c = cp[3];
cp[0] = 1;
}
But this would partially defeat the purpose of using the correct type.
One has to remember to use the address-of operator when assigning an array's address to a pointer-to-array:
char a[10];
char (*p)[10] = &a;
The address-of operator gets the address of the whole array in &a, with the correct type to assign it to p. Without the operator, a is automatically converted to the address of the first element of the array, same as in &a[0], which has a different type.
Since this automatic conversion is already taking place, I am always puzzled that the & is necessary. It is consistent with the use of & on variables of other types, but I have to remember that an array is special and that I need the & to get the correct type of address, even though the address value is the same.
One reason for my problem may be that I learned K&R C back in the 80s, which did not allow using the & operator on whole arrays yet (although some compilers ignored that or tolerated the syntax). Which, by the way, may be another reason why pointers-to-arrays have a hard time to get adopted: they only work properly since ANSI C, and the & operator limitation may have been another reason to deem them too awkward.
When typedef is not used to create a type for the pointer-to-array (in a common header file), then a global pointer-to-array needs a more complicated extern declaration to share it across files:
fileA:
char (*p)[10];
fileB:
extern char (*p)[10];
Well, simply put, C doesn't do things that way. An array of type T is passed around as a pointer to the first T in the array, and that's all you get.
This allows for some cool and elegant algorithms, such as looping through the array with expressions like
*dst++ = *src++
The downside is that management of the size is up to you. Unfortunately, failure to do this conscientiously has also led to millions of bugs in C coding, and/or opportunities for malevolent exploitation.
What comes close to what you ask in C is to pass around a struct (by value) or a pointer to one (by reference). As long as the same struct type is used on both sides of this operation, both the code that hand out the reference and the code that uses it are in agreement about the size of the data being handled.
Your struct can contain whatever data you want; it could contain your array of a well-defined size.
Still, nothing prevents you or an incompetent or malevolent coder from using casts to fool the compiler into treating your struct as one of a different size. The almost unshackled ability to do this kind of thing is a part of C's design.
You can declare an array of characters a number of ways:
char p[10];
char* p = (char*)malloc(10 * sizeof(char));
The prototype to a function that takes an array by value is:
void foo(char* p); //cannot modify p
or by reference:
void foo(char** p); //can modify p, derefernce by *p[0] = 'f';
or by array syntax:
void foo(char p[]); //same as char*
I would not recommend this solution
typedef int Vector3d[3];
since it obscures the fact that Vector3D has a type that you
must know about. Programmers usually dont expect variables of the
same type to have different sizes. Consider :
void foo(Vector3d a) {
Vector3d b;
}
where sizeof a != sizeof b
Maybe I'm missing something, but... since arrays are constant pointers, basically that means that there's no point in passing around pointers to them.
Couldn't you just use void foo(char p[10], int plen); ?
type (*)[];
// points to an array e.g
int (*ptr)[5];
// points to an 5 integer array
// gets the address of the array
type *[];
// points to an array of pointers e.g
int* ptr[5]
// point to an array of five integer pointers
// point to 5 adresses.
On my compiler (vs2008) it treats char (*p)[10] as an array of character pointers, as if there was no parentheses, even if I compile as a C file. Is compiler support for this "variable"? If so that is a major reason not to use it.