Allocating memory for triple pointer - c++

The function somefunction() takes a triple pointer as an argument.
int somefunction(tchar ***returnErrors);
How to allocate memory for the returnErrors parameter?

At a guess . . .
You can think of returnErrors as a pointer to an array of strings.
The first * imples pointer to array
of tchar (or a single string of
tchars)
The second * imples a pointer to an
array of strings.
The last * is so you can change
returnErrors and pass back the new
memory.
To delare memory for this (silly example, allocating memory inside SomeFunction)
tchar ** errors;
// Oops it appears I need to pass back two error strings (+ 1 for null on end, so we know there are no more - thanks tlholaday)
errors = malloc(sizeof(tchar*) * 3);
// the first string has length 20 (+ 1 for null terminator)
errors[0] = malloc(sizeof(tchar) * 21);
// the second string has length 30 (+ 1 for null terminator)
errors[1] = malloc(sizeof(tchar) * 31);
// ensure the last is null
errors[2] = 0;
*returnErrors = errors;
NB: The calling function needs to know that SomeFunction has allocated memory and needs to free it.

Are you implementing somefunction or calling somefunction?
If you are calling somefunction, it is likely that somefunction will be allocating the memory, so all you need to do is pass it a safe place to scribble and clean up afterwards.
tchar **theErrors = 0; // a vector of tchar vectors.
somefunction(&theErrors);
if (theErrors) {
// use the error values
// free the memory somehow - this is for a null-terminated convention
tchar **victim = theErrors;
while (*victim) delete[](*victim++);
delete[] theErrors;
}
Note: I'm using 0 and delete[] instead of NULL and free because the tag says c++.

It depends on what "somefunction" is expecting. You have to investigate on this !
It may be expecting a pointer to a fixed-size array of dimension 2, or a regular array of dimension 3, or ???
In the cases i mentionned, code could look like
tchar errors[SIZE1][SIZE2];
somefunction( &errors );
or
tchar errors[SIZE1][SIZE2][SIZE3];
somefunction( errors );

Does anyone know how to allocate the
memory for the returnErrors parameter?
The question is too generic and cannot be answered in the general case. Following is just an example of a possible code snippet that calls it.
tchar foo;
tchar * p_foo = &foo;
tchar ** pp_foo = &p_foo;
tchar *** ppp_foo = &pp_foo;
somefunction(ppp_foo);
Just a comment: I would consider your function signature unsafe and thus a code smell even if it had one star less.
Also, note:
A pointer is never an array. It is a
variable which contains the value of
a memory address, or NULL.
The address that a pointer contains
doesn't always correspond to the
starting address of an array. int **
p is not always refering to the
starting address of an int[][].
A pointer whose value contains the
starting address of an array is not
the best way to pass this array as a
function parameter. Instead, a reference to the array type can be used.
An array is usually not the best way to contain a set of related values in C++. std::vector and other STL containers should be considered. (However your question has two languages, C and C++, as tags; of course this only applies to the latter possibility. Why the two tags?)

There are two use cases for a funciton like your someFunction() that I would use.
For initialization:
{
tchar **returnErrors;
initErrors(tchar &returnErrors);
/* now returnErrors has a place in memory*/
}
int initErrors(tchar ***returnErrors)
{
*returnErrors = malloc(sizeof(tchar *) * SIZE1)
for (i = 0; i < NUM_ELEMENTS; i++)
(*returnErrors)[i] = malloc(sizeof(tchar) * SIZE2);
/*add checks for malloc failures*/
}
Array passing to a function:
{
returnErrors[SIZE1][SIZE2][SIZE3];
someFunciton(returnErrors);
}
int someFunciton(tchar ***returnErrors)
{
/*assuming that this is a valid index*/
tchar x = returnErrors[1][1][1];
/*return errors is used as a triple array*/
}

The triple pointer does not make any sense if you don't already have a construct of that type. I would rather recommend using a class - or at least a standard container. But if you must know, it's as simple as
tchar ***pointer = (tchar***) malloc(sizeof(tchar**) * amount);

I'd tend agree with tlholaday's guess, with the modification that the function likely returns the number of errors it allocates.
tchar **theErrors = 0; // a vector of tchar vectors.
int nErrors = somefunction(&theErrors);
if (nErrors > 0) {
for (int i = 0; i < nErrors; ++i)
{
printf(theErrors[i]);
free(theErrors[i]);
}
free(theErrors);
}
Note that whether you use free or delete[] will depend on how the memory was allocated in the first place.

Related

Strcpy behavior with stack array c++

Here is my program :
#include <cstring>
const int SIZE =10;
int main()
{
char aName [SIZE]; // creates an array on the stack
std::strcpy(aName, "Mary");
return 0;
}
This program is obviously useless, I am just trying to understand the behavior of the strcpy function.
Here is it's signature :
char * strcpy ( char * destination, const char * source )
so when I do :
std::strcpy(aName, "Mary");
I am passing by value the variable aName. I know that the aName (in the main) contains the address of the array.
So is this assertion correct : strcpy creates a local variable called destination that has as value the address of the array aName that I have created on the stack in the main function?
I am asking this because it is very confusing to me. Whenever I have encountered addresses it usually was to point to a memory allocated on the heap...
Thanks!
Whenever you encounter addresses it doesn't mean it will always point to memory allocated to heap.
You can assign the address of a variable to a pointer like this
int a=5;
int *myPtr= &a;
Now, myPtr is a pointer of type integer which points to the memory of variable which is created on stack which is a have value 5.
So, whenever you create a pointer and assign the (address of) memory using new keyword, it will allocate the memory on heap. So, if I assign the value like this it will be on stack
int *myPtr= new int[5];
So is this assertion correct : strcpy creates a local variable called destination that has as value the address of the array aName that I have created on the stack in the main function?
Yes.
Whenever I have encountered addresses it usually was to point to a memory allocated on the heap...
Yep, usually. But not always.
Pointers to non-dynamically-allocated things are fairly rare in C++, though in C it's more common as that's the only way to have "out arguments" (C does not have references).
strcpy is a function from C's standard library.
Maybe it would help to look at an example implementation of strcpy():
char* strcpy(char* d, const char* s)
{
char* tmp = d;
while (*tmp++ = *s++)
;
return d;
}
That's really all there is to it. Copy characters from the source to the destination until the source character is null (including the null). Return the pointer to the beginning of the destination. Done.
Pointers point to memory. It doesn't matter if that memory is "stack", "heap" or "static".
Function parameters are its local variables.
In this call
std::strcpy(aName, "Mary");
the two arrays (one that is created in main with the automatic storage duration and other is the string literal that has the static storage duration) are implicitly converted to pointers to their first elements.
So you may imagine this call and the function definition the following way
std::strcpy(aName, "Mary");
// …
char * strcpy ( /* char * destination, const char * source */ )
{
char *destination = aName;
const char *source = "Mary";
// …
return destination;
}
Or even like
char *p_to_aName = &aName[0];
const char *p_to_literal = &"Mary"[0];
std::strcpy( p_to_aName, p_to_literal );
// …
char * strcpy ( /* char * destination, const char * source */ )
{
char *destination = p_to_aName;
const char *source = p_to_literal;
// …
return destination;
}
That is within the function its parameters are local variable of pointer types with the automatic storage duration that are initialized by pointers to first characters of the passed character arrays
So is this assertion correct : strcpy creates a local variable called destination that has as value the address of the array aName that I have created on the stack in the main function?
Yes. That is correct. Though I probably wouldn't call it a local variable. It is a parameter. Local variable usually means something like this:
int localVariable;
The word'parameter" is often associated with things like this:
int myFunction(int parameter) {
// use parameter some where...
}
The point is roughly the same though: it creates a variable that will go out of scope once the function exits.
I am asking this because it is very confusing to me. Whenever I have encountered addresses it usually was to point to a memory allocated on the heap...
Yes, this is the most common use case for them. But it isn't their only use. Pointers are addresses, and every variable has an address in memory regardless of whether it is allocated on the "heap" or "stack."
The use here probably because pointers to a char are commonly used to store strings, particularly on older compilers. That combined with the fact that arrays "decay" into pointers, it is probably easier to work with pointers. It is also certainly more backwards compatible to do it this way.
The function could have just as easily used an array, like this:
char * strcpy ( char destination[], const char source[ )
But I'm going to assume it is easier to work with pointers here instead (Note: I don't think you can return an array in C++, so I'm still using char *. However, even if you could, I would imagine it is still easier to work with pointers anyway, so I don't think it makes a lot of difference here.).
Another common use of pointers is using them as a way to sort of "pass by reference":
void foo(int * myX) {
*myX = 4;
}
int main() {
int x = 0;
foo(&x);
std::cout << x; // prints "4"
return 0;
}
However, in modern C++, actually passing by reference is preferred to this:
void foo(int & myX) {
myX = 4;
}
int main() {
int x = 0;
foo(x);
std::cout << x; // prints "4"
return 0;
}
But I bring it up as another example to help drive the point home: memory allocated on the heap isn't the only use of pointers, merely the most common one (though actually dynamically allocated memory has been mostly replaced in modern C++ by things like std::vector, but that is beside the point here).
I know that the aName (in the main) contains the address of the array.
You knew wrong. aName is an array. It contains the elements, not an address.
But when you use the name of the array as a value such as when passing it to strcpy, it is implicitly converted to a pointer to first element of the array (the value of a pointer is the memory address of the pointed object). Such implicit conversion is called decaying.
So is this assertion correct : strcpy creates a local variable called destination that has as value the address of the array aName that I have created on the stack in the main function?
This is correct enough. To clarify: It is a function argument rather than a local variable. But the distinction is not important here. Technically, it is the caller who is responsible for pushing the arguments onto the stack or storing them into registers, so it could be considered that main "creates" the variable.
Whenever I have encountered addresses it usually was to point to a memory allocated on the heap
Pointers are not uniquely associated with "heap". Pretty much any object can be pointed at, whether it has dynamic, static or automatic storage or even if it is a subobject.

Why don't my simple code work in C++?

Error in function realloc(): invalid pointer
int indZero = 0;
int *perZero=NULL;
int zero = 0;//Initialization
ProcessBit(zero,&indZero,&perZero);// Call function
void ProcessBit(int num,int *ind,int **mas)
{
mas=(int**)realloc(&mas,((*ind))*sizeof(int));// Error
mas[num-1]++;//it's correct line
}
A few problems:
The first argument to realloc is the original pointer (or NULL).
Your ProcessBit doesn't really emulate pass-by-reference correctly.
You can use a negative index.
mas is a pointer to a pointer to int, but you use it as a pointer to int.
A "fixed" version might look something like this:
void ProcessBit(int num, int *ind, int **mas)
{
int *temp = realloc(*mas, (*ind + 1) * sizeof(int));
if (temp == NULL)
{
// TODO: Handle error
// TODO: return or exit(EXIT_FAILURE)
}
*mas = temp;
(*mas)[*ind] = 0; // Initial initialization
if (num > 0)
{
(*mas)[num - 1]++;
}
++*ind; // Increase the size
}
Now, if this really was C++ (as you tagged your question) then you should be using std::vector instead, which would solve almost all your problems with much simpler code.
The parameters are wrong. Since you are trying to realloc a NULL pointer it should behave like malloc; however, the header declared in cstdlib is
void* realloc( void* ptr, std::size_t new_size );
The formal parameter mas is already the address of the pointer, so the call should be
*mas=(int*)realloc(*mas,((*ind))*sizeof(int));
(*mas)[num-1]++;
Since realloc handles and returns pointers by copy, not by reference.
You are passing the address of the memory location where the address of a memory location (NULL) is stored to your ProcessBit function, and then the address of that location to the realloc function. The function tries to reallocate memory where the variable mac is stored, on the stack. No wonder it's an invalid pointer.
By passing &mac you are simply taking a step in the wrong direction while dereferencing pointers.

Example why someone should use triple-pointers in C/C++?

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).

C++ new * char aren't empty

I had one question.
I developing server in ASIO and packets are in pointed char.
When i create new char (ex. char * buffer = new char[128];) i must clean it manually to nulls.
By:
for(int i =0;i<128;i++)
{
buffer[i] = 0x00;
}
I doing something wrong, that char isn't clear ?
You do not have to loop over an array of un-initialized values. You can dynamically instantiate array of zeros like this:
char * buffer = new char[128](); // all elements set to 0
^^
There are two types of ways of calling the new operator in C++ - default initialised and zero initialised.
To default initialise (which will leave the value undefined) call:
int * i = new int;
It is then undefined behavoir to read or use this value until its been set.
To zeroinitialise (which will set to 0) use:
int * i = new int();
This also works with arrays:
int * i = new int[4]; // ints are not initialised, you must write to them before reading them
int * i = new int[4](); // ints all zero initialised
There's some more info here
Allocated memory will not be clear, it will contain random stuff instead. That's how memory allocation works. You have to either run a for-loop or use memset to clear it manually.
You also can use calloc. It initializes each elem to 0 automaticaly.
e.g:
char* buffer = (char *) calloc (128, sizeof (char))
First param is number of blocks to be allocated. Second is size of block.
This function returns void* so you have to convert its value to (char *)
If you use calloc (or malloc or any "pure c" allocation functions) you'd better use free function to deallocate memory instead of delete.

C++: sizeof for array length

Let's say I have a macro called LengthOf(array):
sizeof array / sizeof array[0]
When I make a new array of size 23, shouldn't I get 23 back for LengthOf?
WCHAR* str = new WCHAR[23];
str[22] = '\0';
size_t len = LengthOf(str); // len == 4
Why does len == 4?
UPDATE: I made a typo, it's a WCHAR*, not a WCHAR**.
Because str here is a pointer to a pointer, not an array.
This is one of the fine differences between pointers and arrays: in this case, your pointer is on the stack, pointing to the array of 23 characters that has been allocated elsewhere (presumably the heap).
WCHAR** str = new WCHAR[23];
First of all, this shouldn't even compile -- it tries to assign a pointer to WCHAR to a pointer to pointer to WCHAR. The compiler should reject the code based on this mismatch.
Second, one of the known shortcomings of the sizeof(array)/sizeof(array[0]) macro is that it can and will fail completely when applied to a pointer instead of a real array. In C++, you can use a template to get code like this rejected:
#include <iostream>
template <class T, size_t N>
size_t size(T (&x)[N]) {
return N;
}
int main() {
int a[4];
int *b;
b = ::new int[20];
std::cout << size(a); // compiles and prints '4'
// std::cout << size(b); // uncomment this, and the code won't compile.
return 0;
}
As others have pointed out, the macro fails to work properly if a pointer is passed to it instead of an actual array. Unfortunately, because pointers and arrays evaluate similarly in most expressions, the compiler isn't able to let you know there's a problem unless you make you macro somewhat more complex.
For a C++ version of the macro that's typesafe (will generate an error if you pass a pointer rather than an array type), see:
Compile time sizeof_array without using a macro
It wouldn't exactly 'fix' your problem, but it would let you know that you're doing something wrong.
For a macro that works in C and is somewhat safer (many pointers will diagnose as an error, but some will pass through without error - including yours, unfortunately):
Is there a standard function in C that would return the length of an array?
Of course, using the power of #ifdef __cplusplus you can have both in a general purpose header and have the compiler select the safer one for C++ builds and the C-compatible one when C++ isn't in effect.
The problem is that the sizeof operator checks the size of it's argument. The argument passed in your sample code is WCHAR*. So, the sizeof(WCHAR*) is 4. If you had an array, such as WCHAR foo[23], and took sizeof(foo), the type passed is WCHAR[23], essentially, and would yield sizeof(WCHAR) * 23. Effectively at compile type WCHAR* and WCHAR[23] are different types, and while you and I can see that the result of new WCHAR[23] is functionally equivalent to WCHAR[23], in actuality, the return type is WCHAR*, with absolutely no size information.
As a corellary, since sizeof(new WCHAR[23]) equals 4 on your platform, you're obviously dealing with an architecture where a pointer is 4 bytes. If you built this on an x64 platform, you'd find that sizeof(new WCHAR[23]) will return 8.
You wrote:
WCHAR* str = new WCHAR[23];
if 23 is meant to be a static value, (not variable in the entire life of your program) it's better use #define or const than just hardcoding 23.
#define STR_LENGTH 23
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = (size_t) STR_LENGTH;
or C++ version
const int STR_LENGTH = 23;
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = static_cast<size_t>(STR_LENGTH);