I'm searching for an example or explanation why someone should (or should not) use triple-pointers in C/C++.
Are there any examples where triple-pointer arise?
I am especially looking for source-code which uses triple-pointers.
The best example that comes to mind is a sparse multi-level table. For instance one way to implement properties for Unicode characters might be:
prop_type ***proptable;
...
prop_type prop = proptable[c>>14][c>>7&0x7f][c&0x7f];
In this case proptable would need to have a triple-pointer type (and possibly quadruple pointer if the final resulting type is a pointer type). The reason for doing this as multiple levels rather than one flat table is that, at the first and second levels, multiple entries can point to the same subtable when the contents are all the same (e.g. huge CJK ranges).
Here's another example of a multi-level table that I implemented; I can't say I'm terribly proud of the design but given the constraints the code has to satisfy, it's one of the least-bad implementation choices:
http://git.musl-libc.org/cgit/musl/tree/src/aio/aio.c?id=56fbaa3bbe73f12af2bfbbcf2adb196e6f9fe264
If you need to return an array of pointers to variable length strings via a function parameter:
int array_of_strings(int *num_strings, char ***string_data)
{
int n = 32;
char **pointers = malloc(n * sizeof(*pointers));
if (pointers == 0)
return -1; // Failure
char line[256];
int i;
for (i = 0; i < n && fgets(line, sizeof(line), stdin) != 0; i++)
{
size_t len = strlen(line);
if (line[len-1] == '\n')
line[len-1] = '\0';
pointers[i] = strdup(line);
if (pointers[i] == 0)
{
// Release already allocated resources
for (int j = 0; j < i; j++)
free(pointers[j]);
free(pointers);
return -1; // Failure
}
}
*num_strings = i;
*string_data = pointers;
return 0; // Success
}
Compiled code.
If you use a linked list you have to store the address of the first element of the list ( first pointer ) .
If you need to change in that list you need another pointer ( two pointer)
If you need to pass your list that you are changing in two pointers and change it in another function you need another pointer ( three pointer )...
They are a lots of examples
I've used triple pointers in C++:
There is an interface written for a Java program:
https://github.com/BenLand100/SMART/blob/master/src/SMARTPlugin.h
and it takes an array of strings.
typedef void (*_SMARTPluginInit)(SMARTInfo *ptr, bool *replace, int *buttonc, char ***buttonv, int **buttonid, _SMARTButtonPressed *buttonproc);
Then in my program I do:
char* btnTexts[2] = {"Disable OpenGL_Enable OpenGL", "Enable Debug_Disable glDebug"}; //array of C-style strings.
void SMARTPluginInit(SMARTInfo* ptr, bool* ReplaceButtons, int* ButtonCount, char*** ButtonTexts, int** ButtonIDs, _SMARTButtonPressed* ButtonCallback)
{
*ButtonText = btnTexts; //return an array of strings.
}
but in C++, you can use a reference instead of pointer and it'd become:
void SMARTPluginInit(SMARTInfo* ptr, bool* ReplaceButtons, int* ButtonCount, char** &ButtonTexts, int** ButtonIDs, _SMARTButtonPressed* ButtonCallback)
{
ButtonText = btnTexts; //return an array of strings.
}
Notice now that "ButtonTexts" is a reference to an array of C-style strings now.
A char*** can be a pointer to an array of C-style strings and that's one time that you'd use it.
A very simple example is a pointer to an array of arrays of arrays.
Triple pointer is a pointer variable that points to a pointer which in turn points to another pointer. The use of this complex programming technique is that usually in which companies process tons and tons of data at one time .A single pointer would point to a single block of data (suppose in a large file) using the triple pointer would result in 3 times faster processing as different blocks of data(in the same file) can be pointed by different pointer and thus data could be accessed/processed faster (unlike 1 pointer going through the whole file).
Related
I am learning C++ using C++ Primer 5th edition. In particular, i read about void*. There it is written that:
We cannot use a void* to operate on the object it addresses—we don’t know that object’s type, and the type determines what operations we can perform on that object.
void*: Pointer type that can point to any nonconst type. Such pointers may not
be dereferenced.
My question is that if we're not allowed to use a void* to operate on the object it addressess then why do we need a void*. Also, i am not sure if the above quoted statement from C++ Primer is technically correct because i am not able to understand what it is conveying. Maybe some examples can help me understand what the author meant when he said that "we cannot use a void* to operate on the object it addresses". So can someone please provide some example to clarify what the author meant and whether he is correct or incorrect in saying the above statement.
My question is that if we're not allowed to use a void* to operate on the object it addressess then why do we need a void*
It's indeed quite rare to need void* in C++. It's more common in C.
But where it's useful is type-erasure. For example, try to store an object of any type in a variable, determining the type at runtime. You'll find that hiding the type becomes essential to achieve that task.
What you may be missing is that it is possible to convert the void* back to the typed pointer afterwards (or in special cases, you can reinterpret as another pointer type), which allows you to operate on the object.
Maybe some examples can help me understand what the author meant when he said that "we cannot use a void* to operate on the object it addresses"
Example:
int i;
int* int_ptr = &i;
void* void_ptr = &i;
*int_ptr = 42; // OK
*void_ptr = 42; // ill-formed
As the example demonstrates, we cannot modify the pointed int object through the pointer to void.
so since a void* has no size(as written in the answer by PMF)
Their answer is misleading or you've misunderstood. The pointer has a size. But since there is no information about the type of the pointed object, the size of the pointed object is unknown. In a way, that's part of why it can point to an object of any size.
so how can a int* on the right hand side be implicitly converted to a void*
All pointers to objects can implicitly be converted to void* because the language rules say so.
Yes, the author is right.
A pointer of type void* cannot be dereferenced, because it has no size1. The compiler would not know how much data he needs to get from that address if you try to access it:
void* myData = std::malloc(1000); // Allocate some memory (note that the return type of malloc() is void*)
int value = *myData; // Error, can't dereference
int field = myData->myField; // Error, a void pointer obviously has no fields
The first example fails because the compiler doesn't know how much data to get. We need to tell it the size of the data to get:
int value = *(int*)myData; // Now fine, we have casted the pointer to int*
int value = *(char*)myData; // Fine too, but NOT the same as above!
or, to be more in the C++-world:
int value = *static_cast<int*>(myData);
int value = *static_cast<char*>(myData);
The two examples return a different result, because the first gets an integer (32 bit on most systems) from the target address, while the second only gets a single byte and then moves that to a larger variable.
The reason why the use of void* is sometimes still useful is when the type of data doesn't matter much, like when just copying stuff around. Methods such as memset or memcpy take void* parameters, since they don't care about the actual structure of the data (but they need to be given the size explicitly). When working in C++ (as opposed to C) you'll not use these very often, though.
1 "No size" applies to the size of the destination object, not the size of the variable containing the pointer. sizeof(void*) is perfectly valid and returns, the size of a pointer variable. This is always equal to any other pointer size, so sizeof(void*)==sizeof(int*)==sizeof(MyClass*) is always true (for 99% of today's compilers at least). The type of the pointer however defines the size of the element it points to. And that is required for the compiler so he knows how much data he needs to get, or, when used with + or -, how much to add or subtract to get the address of the next or previous elements.
void * is basically a catch-all type. Any pointer type can be implicitly cast to void * without getting any errors. As such, it is mostly used in low level data manipulations, where all that matters is the data that some memory block contains, rather than what the data represents. On the flip side, when you have a void * pointer, it is impossible to determine directly which type it was originally. That's why you can't operate on the object it addresses.
if we try something like
typedef struct foo {
int key;
int value;
} t_foo;
void try_fill_with_zero(void *destination) {
destination->key = 0;
destination->value = 0;
}
int main() {
t_foo *foo_instance = malloc(sizeof(t_foo));
try_fill_with_zero(foo_instance, sizeof(t_foo));
}
we will get a compilation error because it is impossible to determine what type void *destination was, as soon as the address gets into try_fill_with_zero. That's an example of being unable to "use a void* to operate on the object it addresses"
Typically you will see something like this:
typedef struct foo {
int key;
int value;
} t_foo;
void init_with_zero(void *destination, size_t bytes) {
unsigned char *to_fill = (unsigned char *)destination;
for (int i = 0; i < bytes; i++) {
to_fill[i] = 0;
}
}
int main() {
t_foo *foo_instance = malloc(sizeof(t_foo));
int test_int;
init_with_zero(foo_instance, sizeof(t_foo));
init_with_zero(&test_int, sizeof(int));
}
Here we can operate on the memory that we pass to init_with_zero represented as bytes.
You can think of void * as representing missing knowledge about the associated type of the data at this address. You may still cast it to something else and then dereference it, if you know what is behind it. Example:
int n = 5;
void * p = (void *) &n;
At this point, p we have lost the type information for p and thus, the compiler does not know what to do with it. But if you know this p is an address to an integer, then you can use that information:
int * q = (int *) p;
int m = *q;
And m will be equal to n.
void is not a type like any other. There is no object of type void. Hence, there exists no way of operating on such pointers.
This is one of my favourite kind of questions because at first I was also so confused about void pointers.
Like the rest of the Answers above void * refers to a generic type of data.
Being a void pointer you must understand that it only holds the address of some kind of data or object.
No other information about the object itself, at first you are asking yourself why do you even need this if it's only able to hold an address. That's because you can still cast your pointer to a more specific kind of data, and that's the real power.
Making generic functions that works with all kind of data.
And to be more clear let's say you want to implement generic sorting algorithm.
The sorting algorithm has basically 2 steps:
The algorithm itself.
The comparation between the objects.
Here we will also talk about pointer functions.
Let's take for example qsort built in function
void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void*))
We see that it takes the next parameters:
base − This is the pointer to the first element of the array to be sorted.
nitems − This is the number of elements in the array pointed by base.
size − This is the size in bytes of each element in the array.
compar − This is the function that compares two elements.
And based on the article that I referenced above we can do something like this:
int values[] = { 88, 56, 100, 2, 25 };
int cmpfunc (const void * a, const void * b) {
return ( *(int*)a - *(int*)b );
}
int main () {
int n;
printf("Before sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
qsort(values, 5, sizeof(int), cmpfunc);
printf("\nAfter sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
return(0);
}
Where you can define your own custom compare function that can match any kind of data, there can be even a more complex data structure like a class instance of some kind of object you just define. Let's say a Person class, that has a field age and you want to sort all Persons by age.
And that's one example where you can use void * , you can abstract this and create other use cases based on this example.
It is true that is a C example, but I think, being something that appeared in C can make more sense of the real usage of void *. If you can understand what you can do with void * you are good to go.
For C++ you can also check templates, templates can let you achieve a generic type for your functions / objects.
#include <iostream>
char** make2D(const int dim1, const int dim2)
{
char* toAlloc;
const int size = (dim1 * dim2) + dim2;
toAlloc = new char[size];
for(int i = 0; i < dim2; i++)
{
toAlloc[i] = reinterpret_cast<char>(&toAlloc[(dim2 + (dim1 * i))]);
}
return reinterpret_cast<char**>(toAlloc);
}
int main(void)
{
int dim1 = 8;
int dim2 = 10;
char** array2D = make2D(dim1, dim2);
for (int i = 0; i < dim2; ++i)
{
array2D[i][i % dim1] = i + 100; // << Crash
}
return 0;
}
I was trying to allocate two dimensional array by a single allocation.
So, my algorithm was, first 10(which is dim2 in this code) items has pointer to first item of each rows.
When I was try this by pointer to 'int',
int** make2D(const int dim1, const int dim2)
{
int* toAlloc;
const int size = (dim1 * dim2) + dim2;
toAlloc = new int[size];
for(int i = 0; i < dim2; i++)
{
toAlloc[i] = reinterpret_cast<int>(&toAlloc[(dim2 + (dim1 * i))]);
}
return reinterpret_cast<int**>(toAlloc);
}
int main(void)
{
int dim1 = 8;
int dim2 = 10;
int** array2D = make2D(dim1, dim2);
for (int i = 0; i < dim2; ++i)
{
array2D[i][i % dim1] = i + 100;
}
return 0;
}
it works fine but when I do this in char, it crashes in commented line in above code.
My thought of crashing was when I do reinterpret_cast, something happens because of memory size gap between pointer(8byte) and char(1byte).
So like, sounds like ridiculous... changing pointer(8byte) to int(4byte) was fine, but when I do cast more dramatically(8byte to 1byte), it causes some problems...
I have no idea why char doesn't work but int works.
Could you give some advice to make char case works?
To answer the question yes there is a difference, a huge one, on many platforms a pointer might fit into an int, on very few platforms it will fit into a char. On modern PCs which are 64-bit none are safe ways to store a pointer.
Use containers such as vector or array if the size is static.
Try something like:
array<array<T, dim2>, dim1> variable{};
if you actually want a 2-dimensional array of type T; since you seem to need an array of pointers try something like:
array<array<T *, dim2>, dim1> variable{};
This will take care to make an array of the appropriate type to store pointers in for your platform, no matter how big pointers actually are, obviously you should replace T with the proper type of the data you want to point to, this will ensure pointer math is done properly for you.
Array types will have their size calculated at compile time, if you need dynamic sizes you should use vector, after allocation call resize on the vector and all sub-vectors to make sure you allocate all the memory in as few passes as possible.
Please also don't use reinterpret_cast, or c-style casts, it's a recipe for disaster unless you know very well what you're doing.
Don't know what book you're reading or who is teaching you C++ but please change your knowledge source.
Using raw owning pointers is discouraged and the way you're using them is wrong in so many ways.
Never store a pointer in anything but a pointer type. Even in plain C you should cast to at least void * if you need to cast at all.
Please read about unique_ptr or shared_ptr if you really want to store/pass pointers around directly.
If you insist on using raw pointers for containers please try building your code with sanitizers such as address sanitizer, memory sanitizer (these are supported at least by clang and gcc, possibly more compilers these days)
Issue is "incompatibility" of size of objects:
sizeof(char) is 1
sizeof(int) is generally 4 or 8 (but at least 2).
sizeof(T*) which is generally 4 or 8, std::uintp_t can hold void* value, which is not necessary the case with int (and even less with char).
You cannot store safely void* into char or int. It happens it's working for you for int, but it is not portable.
reinterpret_cast is generally the wrong tool.
Simpler would be to create a class Matrix, with std::vector<T> and accessor to fix indexing. (You might even have proxy to allow m[2][3] syntax).
With owning raw pointer, you need need placements new, and provide correct deleter...
I have an array of bytes that I want to change how it's accessed. Here's the array:
char bytes[100];
I want to have another array that changes how the original array is accessed. If we could put references in array, it would look something like this:
char& bytes_ref[50];
for(size_t i = 0; i < 50; i++){
bytes_ref[i] = bytes[i * 2];
}
Though array of references aren't valid. I came up with this that in theory does what I want:
#include <iostream>
struct Byte {
char* ref;
Byte(){}
Byte(char& c){
ref = &c;
}
operator char&() const {
return *ref;
}
};
int main(){
char bytes[100];
Byte bytes_ref[50];
for(size_t i = 0; i < 50; i++){
bytes_ref[i] = bytes[i * 2];
}
}
Though this works like I want it to, it takes up a ton of space.
Is there some way to do this without using up 7 more bytes per entry? My instinct is no, but I'm hopeful there may be some kind of special functionality in C++ that could do this, or maybe do some direct memory manipulation and access.
The reason I want to do this is I have an array of bytes that represent an image in NV21 format. I want to create a separate array that references the bytes in the original buffer but takes half the pixels, effectively resizing the image on the fly.
I have to pass a char** to an uncontrolled library.
Thanks in advance!
I do not think you can use less memory than one pointer per element
If you need to pass an array of char* to an uncontrolled library, I understand that you have to allocate the whole array because that is what the library is expecting. The code would like:
std::array<char*, 50> bytes_ref;
for(size_t i = 0; i < bytes_ref.size(); ++i){
bytes_ref[i] = &bytes[i * 2];
}
Then, you can pass the array as:
f(bytes_ref.data());
If you would not need to use char**, you would have more options. You may use std::reference_wrapper although its size is similar to a pointer. If you always have to skip 2 elements, you may create an object for this. For example:
class Wrapper {
public:
Wrapper(char *origin) : m_origin(origin) {}
char* operator[](size_t index) {return m_origin + 2*index;}
private:
char *m_origin;
};
The above will return a pointer to every 2 elements using the [] operator.
[edit] Outside of this get method (see below), i'd like to have a pointer double * result; and then call the get method, i.e.
// Pull results out
int story = 3;
double * data;
int len;
m_Scene->GetSectionStoryGrid_m(story, data, len);
with that said, I want to a get method that simply sets the result (*&data) by reference, and does not dynamically allocate memory.
The results I am looking for already exist in memory, but they are within C-structs and are not in one continuous block of memory. Fyi, &len is just the length of the array. I want one big array that holds all of the results.
Since the actual results that I am looking for are stored within the native C-struct pointer story_ptr->int_hv[i].ab.center.x;. How would I avoid dynamically allocating memory like I am doing above? I’d like to point the data* to the results, but I just don’t know how to do it. It’s probably something simple I am overlooking… The code is below.
Is this even possible? From what I've read, it is not, but as my username implies, I'm not a software developer. Thanks to all who have replied so far by the way!
Here is a snippet of code:
void GetSectionStoryGrid_m( int story_number, double *&data, int &len )
{
std::stringstream LogMessage;
if (!ValidateStoryNumber(story_number))
{
data = NULL;
len = -1;
}
else
{
// Check to see if we already retrieved this result
if ( m_dStoryNum_To_GridMap_m.find(story_number) == m_dStoryNum_To_GridMap_m.end() )
{
data = new double[GetSectionNumInternalHazardVolumes()*3];
len = GetSectionNumInternalHazardVolumes()*3;
Story * story_ptr = m_StoriesInSection.at(story_number-1);
int counter = 0; // counts the current int hv number we are on
for ( int i = 0; i < GetSectionNumInternalHazardVolumes() && story_ptr->int_hv != NULL; i++ )
{
data[0 + counter] = story_ptr->int_hv[i].ab.center.x;
data[1 + counter] = story_ptr->int_hv[i].ab.center.y;
data[2 + counter] = story_ptr->int_hv[i].ab.center.z;
m_dStoryNum_To_GridMap_m.insert( std::pair<int, double*>(story_number,data));
counter += 3;
}
}
else
{
data = m_dStoryNum_To_GridMap_m.find(story_number)->second;
len = GetSectionNumInternalHazardVolumes()*3;
}
}
}
Consider returning a custom accessor class instead of the "double *&data". Depending on your needs that class would look something like this:
class StoryGrid {
public:
StoryGrid(int story_index):m_storyIndex(story_index) {
m_storyPtr = m_StoriesInSection.at(story_index-1);
}
inline int length() { return GetSectionNumInternalHazardVolumes()*3; }
double &operator[](int index) {
int i = index / 3;
int axis = index % 3;
switch(axis){
case 0: return m_storyPtr->int_hv[i].ab.center.x;
case 1: return m_storyPtr->int_hv[i].ab.center.y;
case 2: return m_storyPtr->int_hv[i].ab.center.z;
}
}
};
Sorry for any syntax problems, but you get the idea. Return a reference to this and record this in your map. If done correctly the map with then manage all of the dynamic allocation required.
So you want the allocated array to go "down" in the call stack. You can only achieve this allocating it in the heap, using dynamic allocation. Or creating a static variable, since static variables' lifecycle are not controlled by the call stack.
void GetSectionStoryGrid_m( int story_number, double *&data, int &len )
{
static g_data[DATA_SIZE];
data = g_data;
// continues ...
If you want to "avoid any allocation", the solution by #Speed8ump is your first choice! But then you will not have your double * result; anymore. You will be turning your "offline" solution (calculates the whole array first, then use the array elsewhere) to an "online" solution (calculates values as they are needed). This is a good refactoring to avoid memory allocation.
This answer to this question relies on the lifetime of the doubles you want pointers to. Consider:
// "pointless" because it takes no input and throws away all its work
void pointless_function()
{
double foo = 3.14159;
int j = 0;
for (int i = 0; i < 10; ++i) {
j += i;
}
}
foo exists and has a value inside pointless_function, but ceases to exist as soon as the function exits. Even if you could get a pointer to it, that pointer would be useless outside of pointless_function. It would be a dangling pointer, and dereferencing it would trigger undefined behavior.
On the other hand, you are correct that if you have data in memory (and you can guarantee it will live long enough for whatever you want to do with it), it can be a great idea to get pointers to that data instead of paying the cost to copy it. However, the main way for data to outlive the function that creates it is to call new, new[], or malloc. You really can't get out of that.
Looking at the code you posted, I don't see how you can avoid new[]-ing up the doubles when you create story. But you can then get pointers to those doubles later without needing to call new or new[] again.
I should mention that pointers to data can be used to modify the original data. Often that can lead to hard-to-track-down bugs. So there are times that it's better to pay the price of copying the data (which you're then free to muck with however you want), or to get a pointer-to-const (in this case const double* or double const*, they are equivalent; a pointer-to-const will give you a compiler error if you try to change the data being pointed to). In fact, that's so often the case that the advice should be inverted: "there are a few times when you don't want to copy or get a pointer-to-const; in those cases you must be very careful."
//Prints out a given array
template <typename T>
void print(T t)
{
for(int i = 0; i < t.size(); i++)
{
cout << t[i] << " ";
}
cout << endl;
}
I have an idea but it includes passing the size of the array. Is it possible to avoid this?
*Update
Thanks for all of the answers/ideas but this problem is getting way deeper than my snorkeler can handle. I wanted to rewrite my C++ code in C because it was horribly written and slow. I see now that I have an opportunity to make it even worse in C. I'll rewrite it from the ground up in Python(performance be damned). Thanks again
If you don't have ELEMENTS, it's
#define ELEMENTS(a) (sizeof(a)/sizeof(*a))
Then,
#define print_array(a, specifier) print_array_impl(a, specifier, ELEMENTS(a), sizeof(*a))
void print_array_impl(void* a, char* specifier, size_t asize, size_t elsize)
{
for(int i = 0; i < asize; i++)
{
// corrected based on comment -- unfortunately, not as general
if (strcmp(specifier, "%d") == 0)
printf(specifier, ((int*)a)[i]);
// else if ... // check other specifiers
printf(" ");
}
printf("\n");
}
Use like this
print_array(a, "%d") // if a is a int[]
and, a needs to be an array name, not a pointer (or else ELEMENTS won't work)
You cannot know what is the size of an array without passing the size of that array (except operating with sizeof in static arrays). This is because the a pointer to a block of memory will only point to the base of the block of memory, from which you can know where the array/block of memory starts, but as there is no end defined you cannot determine where it will end.
You either need to set your own length per array and preserve it, and use it with the array like as described:
You can make a new type like:
struct _my_array {
typename arr[MAX];
int n;
} my_array;
OR
struct _my_array {
typename *arr;
int n;
} my_array;
In this case you need to allocate the a block of memory dynamically with new or malloc , and when finished free the memory with delete or free (respectively).
Or you can simply pass the array number of elements through the function.
Another way is to use a special terminator value of your array type which if encountered will be determined as the end of the array. In this case you need not preserve the size. For example a string is '\0' terminated, so all the string functions know that when a '\0' character is encounter in the char array it will consider that the string has end.
UPDATE
Because this is a generic function and the array can be of any type, one thing which you can do is like this:
struct _my_generic_arr {
void *arr;
int n;
int type;
} my_generic_arr;
When populating this array you can use any type. To identify which type, pass an identified in the type component. Each unique value will determine which type does the arr pointer actually points to (was actually the intended type to be pointed). The n will define the length. Now, depending on different values of type make a switch - case or an if - else ladder or nest, and process the array as you need.
It is impossible in c to track the size of an array in other block,,
It would be a better option to pass the size of the array along..
The other option would be to declare a global variable that has the size and using that variable inside the function
Eg,,
int size=<some value>
void main()
{
int arr[<same value>];
}
void print(T t)
{
for(int i = 0; i < size; i++)
{
printf("%d ",t[i]) //assuming T as int
}
printf("\n");
}
In C, you would need to pass two additional parameters: the size of the array (as you mentioned), and some way of indicating how to convert t[i] into a string. To convert t[i] to a string, you could create a custom switch statement to decode possible types, pass a pointer to a function that will return the string pointer, or you could pass the printf format specifier (e.g. "%d" for integer).
The problem is larger than you think. If you have an array of size 12, how do you know what data is held in that array? It could be 3 char*'s (on 32 bit system), 3 int32_t's, or even 12 chars. You have no way of knowing how to interpret the data. The best you could do is to implement your own version of a v-table and putting a print or toString function into it.
typedef struct {
void *array;
size_t length;
int element_width;
printer_t to_string;
} container;
printer_t is a type that describes a function pointer that takes an element pointer and returns a string (or prints it, if you don't want to free the string). This is almost never worth doing in C. That doesn't mean it can't be done. I would emphasize, though, that none of this is intended to imply that it should be done.
The function itself would look something like this:
void print(container *thing)
{
size_t offset;
int width;
char *stringified;
width = thing->element_width;
for (offset = 0; offset * width < thing->length; offset += width)
{
stringified = thing->to_string(thing->array + offset);
printf("%s ", stringified);
free(stringified);
}
}
What this does is essentially turn a struct into a faux class with a function pointer for a method. You could be more object-oriented and put the method in the type being printed and make it an array of those instead. Either way, it's not a good idea. C is for writing C. If you try to write in a different language, you'll end up with all sorts of terrible stuff like this.