I have a large char array which is functioning as a memory pool and want to store a pointer in the first position of the array which points to whatever the next open position in the pool is, so every time something is allocated to the pool the pointer would point to the byte that follows the ones which were just allocated. My only problem is I am not quite sure how to store the pointer in the array and then be able to modify it in other functions since the only place the pointer will exist is in the array[0] position. Can anyone point me in the right direction on this?
the char array is declared like:
char pool[size];
What you really want is an index into that array that tells you where to insert.
You could declare a structure:
struct pool
{
char poolData[size];
int insertIndex;
};
So that you always have the pool memory and index where you want to insert to together. Or, just have a separate variable and pass them together to whoever needs to use it.
char pool[size];
int insertIndex;
There's no need to "hijack" the first element of the array and use it differently than the rest of the array; just declare another variable to track the state of the pool.
If you can't follow the recommendations of the other answers because you absolutely must use the pool to store all information in it, the safest way to store the integer information in the char-array is to use memcpy (I am using C++11 syntax):
#include <cstring>
#include <iostream>
int main()
{
using size_type = std::size_t;
static constexpr size_type size = 1000;
char pool[size];
/* Store 12 as integer information at the beginning
of the pool: */
size_type next_free = 12;
std::memcpy(pool,&next_free,sizeof(size_type));
/* Retrieve it: */
size_type retrieved = 0;
std::memcpy(&retrieved,pool,sizeof(size_type));
/* This will output 12: */
std::cout << retrieved << std::endl;
return 0;
}
Of course this implies that the first sizeof(size_type) entries of the pool must not be used to store any actual characters. The lowest entry you can actually use is pool[sizeof(size_type)].
I think what you need is an index to remember where is the next empty spot on the pool (initially will be zero).
char pool[size];
int index = 0;
Then everytime you insert a new element, you just increment it:
if(index < size) {
pool[index] = 123;
index++;
}
char **pPoolEnd = (char **) pool;
to initialize the pointer you want.
*pPoolEnd = pool + sizeof(char **);
to make it point to its own end (e.g. when there's nothing else in the pool).
However, why would you want to do this? It's confusing, error prone and probably unnecessary. Others have pointed to much better alternatives. Assuming I had such a pool in the first place, I would probably go with one of those, or simply use a separate pointer, char *poolEnd along with pool.
Also, it's bad style to expose your implementation details to users ("pool end pointer is at pool[0]") and even worse to expect them to deal with them ("please update pool[0] whenever you'd like to allocate from the pool"). Think malloc and free; expose simple function interfaces to your users.
Related
I would like to perform a deep copy of a char**, but I have no idea how to allocate memory / copy this datatype. This is for a copy constructor in a class that contains a char**. For example, lets say I have this code:
char ** arr1 = new char*[20];
arr1[0] = (char*)"This is index 1";
arr1[1] = (char*)"This is index 2";
char ** arr2;
How do I deep copy the contents of arr1 into arr2? Any help is appreciated!
It’s for a programming assignment, and the teacher wants all strings
to be stored as char*,...
You can tell your teacher that std::string does store strings as char*. If he still doesnt like you to use std::string you should write your own wrapper, because working with bare char* is what you do when you write C, but not in C++. You should write a:
struct my_string {
char* data;
... constructor, operator[], etc...
};
You basically dont need to write more code than you already do, but you should put it in the right place (ie hide it behind a nice interface). You will immediately see the benefit of it when you eg consider ...
...so an array of strings has to be stored as an array of char*.
No. An array of strings is std::array<my_string> (or std::vector<my_string> if it is supposed to be dynamic). And if your teacher insists on not using std::vector, then you should do the same as you just did for strings for vectors (ie encapsulate all the dirty pointer and memory stuff in one place).
This seems more like a C question, but here is an example:
char **AllocateAndDeepCopy(char **arr1, int arr1size)
{
unsigned int i;
char **arr2;
/* Allocate string array */
arr2 = new char*[arr1size];
/* Iterate array elements */
for (i=0; i<arr1size; i++) {
/* Allocate string */
arr2[i] = new char[strlen(arr1[i])+1];
/* Copy contents */
strcpy(arr2[i], arr1[i]);
}
return arr2;
}
Later you have to deallocate arr2 this way:
void DeallocateArr2(char **arr2, int size)
{
for (int i=0; i<size; i++) {
delete arr2[i];
}
delete arr2;
}
I can only shake my head about the sorry state of C++ education. We have a looong way to go there. But since that’s apparently a given, what’s the best you can do?
To copy a C-style data structure like that you have know two things at the point of copy. Both are not inherently provided by a C-style array, so you’ll have to track them explicitely.
The capacity of arr1: 20. If that’s not a compile time constant you have to store it and pass it around. Since you want to implement a copy ctor that means storing the capacity in a non-static member variable of the object.
The number of used indexes in arr1: 2. Same as above. Alternatively make sure that all unused indexes are set to nullptr.
Now you can allocate an arr2 of the correct size and then allocate+memcpy all used indexes.
However, your program will go up in flames regardless, because arr1 and arr2 cannot be treated the same, even though they look identical. The used indexes of arr1 must never ever be deleted because they contain pointers to character literals: They were never newd and live in read-only memory. On the other hand you absolutely must delete the indexes of arr2, because they were newd.
If this brutal disregard of const is really required by the assignment I’d go one step further. I’d introduce another member variable, an array of booleans that tracks which indexes of the char array point to char literals and which were dynamically allocated. During copy you now have all the necessary information to either memcpy or simply set the pointer. Crazy? Definitely, but the whole assignment is, and that way the craziness is visible at least instead of hidden behind an innocent-looking C-style cast. Btw: those should be const_cast<char*> to make it clear what’s going on.
Just take a look at http://en.cppreference.com/w/cpp/algorithm/copy, the deep copy is made by
*d_first++ = *first++;
#include <iostream>
using namespace std;
void mystrcat(char destination[], const char source[]){
int counter = 0;
while(source[counter] != '/0'){
destination += source[counter];
counter++;
}
destination += '/0';
}
int main(){
}
For my code, when I try to concatenate with the function mystrcat, my school's test bed says that there was a segmentation error. The purpose of my function is to concatenate while removing the NULL from the end of destination and adding it to the end of source. Is there a segmentation error because I am not removing NULL? If so, how do I access the last element of the array? The number of elements is unknown so I don't know if I can use pop_back. Thank you.
Edit:
#include <iostream>
using namespace std;
void mystrcat(char destination[], const char source[]){
int counter = 0;
int counter2 = 0;
while(destination[counter2] != '/0'){
counter2++;
}
while(source[counter] != '/0'){
destination[counter2 - 1] = source[counter];
counter++;
}
destination[counter] = '/0';
}
int main(){
}
is my edited code but now the testbed says that it is taking too long and crashes.
Arrays in C++ have an interesting property where they can easily decay to pointers.
destination += source[counter]; does not append source[counter] to the end of destination.
Instead, destination has decayed to a pointer and this operation is doing pointer arithmatic on destination.
Instead, you want to do destination[destinationLocation] = source[counter]; to actually set the character in destination.
Don't forget to set destination[end] = '\0'; at the end to null terminate the destination array.
One final thing to watch out for is that C++ will not make sure that your arrays are properly sized. If destination is not of the proper size, the code will fail at run time with a segmentation fault.
For future reference, you might want to look into using C++'s std::vector class. std::vector is a variable sized array-like container which automatically keeps track of its size and memory usage. Using pure arrays in C++ is sometimes difficult and error prone (as you have just seen), so std::vector can make things easier.
Static arrays are a fixed sized in C++. That means you can't extend or shrink them at all under any circumstance.
There are three main alternatives to achieve what you're trying to do. From your code, it looks like the best one would be to use the std::string class (since you're using chars). It contains all the code to manage and concatenate the data very easily.
If you absolutely need an array, then std::vector would be the next best choice. It allows push and pop operations etc., and will automatically resize the underlying dynamic array as needed.
Finally, you could use a dynamic array directly. You would create and destroy it using new [] and delete []. When you need to extend the array, you need to create a new one with the new size, copy the old data over, and destroy the old one. It's a lot of unnecessary extra work though, so the other two options are much better.
I have a quick question regarding the scope of dynamic arrays, which I assume is causing a bug in a program I'm writing. This snippet checks a function parameter and branches to either the first or the second, depending on what the user passes.
When I run the program, however, I get a scope related error:
error: ‘Array’ was not declared in this scope
Unless my knowledge of C++ fails me, I know that variables created within a conditional fall out of scope when when the branch is finished. However, I dynamically allocated these arrays, so I cannot understand why I can't manipulate the arrays later in the program, since the pointer should remain.
//Prepare to store integers
if (flag == 1) {
int *Array;
Array = new int[input.length()];
}
//Prepare to store chars
else if (flag == 2) {
char *Array;
Array = new char[input.length()];
}
Can anyone shed some light on this?
Declare Array before if. And you can't declare array of different types as one variable, so I think you should use to pointers.
int *char_array = nullptr;
int *int_array = nullptr;
//Prepare to store integers
if (flag == 1) {
int_array = new int[input.length()];
}
//Prepare to store chars
else if (flag == 2) {
char_array = new char[input.length()];
}
if (char_array)
{
//do something with char_array
}
else if (int_array)
{
//do something with int_array
}
Also as j_random_hacker points, you might want to change you program design to avoid lot's of if
While you are right that since you dynamically allocated them on the heap, the memory won't be released to the system until you explicitly delete it (or the program ends), the pointer to the memory falls out of scope when the block it was declared in exits. Therefore, your pointer(s) need to exist at a wider scope if they will be used after the block.
The memory remains allocated (i.e. taking up valuable space), there's just no way to access it after the closing }, because at that point the program loses the ability to address it. To avoid this, you need to assign the pointer returned by new[] to a pointer variable declared in an outer scope.
As a separate issue, it looks as though you're trying to allocate memory of one of 2 different types. If you want to do this portably, you're obliged to either use a void * to hold the pointer, or (less commonly done) a union type containing a pointer of each type. Either way, you will need to maintain state information that lets the program know which kind of allocation has been made. Usually, wanting to do this is an indication of poor design, because every single access will require switching on this state information.
If I understand your intend correctly what you are trying to do is: depending on some logic allocate memory to store n elements of either int or char and then later in your function access that array as either int or char without the need for a single if statement.
If the above understanding is correct than the simple answer is: "C++ is a strong-typed language and what you want is not possible".
However... C++ is also an extremely powerful and flexible language, so here's what can be done:
Casting. Something like the following:
void * Array;
if(flag1) Array = new int[len]
else Array = new char[len];
// ... later in the function
if(flag) // access as int array
int i = ((int*)Array)[0];
Yes, this is ugly and you'll have to have those ifs sprinkled around the function. So here's an alternative: template
template<class T> T foo(size_t _len)
{
T* Array = new T[_len];
T element = Array[0];
return element;
}
Yet another, even more obscure way of doing things, could be the use of unions:
union int_or_char {int i; char c;};
int_or_char *Array = new int_or_char[len];
if(flag) // access as int
int element = Array[0].i;
But one way or the other (or the third) there's no way around the fact that the compiler has to know how to deal with the data you are trying to work with.
Turix's answer is right. You need to keep in mind that two things are being allocated here, The memory for the array and the memory when the location of the array is stored.
So even though the memory from the array is allocated from the heap and will be available to the code where ever required, the memory where the location of the array is stored (the Array variable) is allocated in the stack and will be lost as soon as it goes out of scope. Which in this case is when the if block end. You can't even use it in the else part of the same if.
Another different code suggestion from Andrew I would give is :
void *Array = nullptr;
if (flag == 1) {
Array = new int[input.length()];
} else if (flag == 2) {
Array = new char[input.length()];
}
Then you can directly use if as you intended.
This part I am not sure : In case you want to know if its an int or char you can use the typeid literal. Doesn't work, at least I can't get it to work.
Alternatively you can use your flag variable to guess what type it is.
I need to use a function (part of an API) which stores some requested data into a dynamic sized struct using call by reference. The struct is defined as follows - it concerns access control lists of either posix or NFS4 version, but that is just the use case, I guess.
typedef struct my_struct
{
unsigned int len; /* total length of the struct in bytes */
... /* some other static sized fields */
unsigned int version; /* 2 different versions are possible */
unsigned int amount; /* number of entries that follow */
union {
entry_v1_t entry_v1[1];
entry_v2_t entry_v2[1];
};
} my_struct_t;
There are 2 versions of the entries and I know which one I will obtain (v1). Both entry_v1_t and entry_v2_t are fixed (but different) sized structs just containing integers (so I guess they are not worth being explained here). Now I need to use an existing function to fill my structure with the information I need using Call by Reference, the signature is as follows, including the comments - I don't have access to the implementation:
int get_information(char *pathname, void *ptr);
/* The ptr parameter must point to a buffer mapped by the my_struct
* structure. The first four bytes of the buffer must contain its total size.
*/
So the point is, that I must allocate memory for that struct but don't know for how much entries (and, as consequence, the total size) I must allocate. Have you ever dealt with such a situation?
Under Windows API there are many such functions, you normally call them with some NULL pointer to get size of the buffer, then call again with allocated buffer. In case during next call size of buffer have changed function returns error and you need allocate again. So you do it in a while loop till function returns with success.
So your get_information must implement somehow such mechanisms, either it returns false if buffer is to small or returns its correct size if ptr is NULL. But that is just my guess.
OK I thing I figured out how it works. Thanks for your ideas and notes. I declared a my_struct pointer and allocated minimum space for the fixed sized fields (5) before the dynamic array => 5 * sizeof(unsigned int). Invoking get_information with that pointer returns -1 and sets errno = 28 and strerror(errno) = "No space left on device".
But, it sets the my_struct->len field to the required size and that seems to be the answer to my question - how should you know? No I can invoke get_information initially with the minimum space and figure out how much I need to allocate, and afterwards call it again with the right sized memory allocated to get the information successfully.
The loop solution seems to make sense anyway and would have been my next try - since there are usually just a few entries in that dynamic array.
Thank you.
I apologise if I'm completely misunderstanding C++ at the moment, so my question might be quite simple to solve. I'm trying to pass a character array into a function by value, create a new array of the same size and fill it with certain elements, and return that array from the function. This is the code I have so far:
char *breedMutation(char genome []){
size_t genes = sizeof(genome);
char * mutation = new char[genes];
for (size_t a = 0 ;a < genes; a++) {
mutation[a] = 'b';
}
return mutation;
}
The for loop is what updates the new array; right now, it's just dummy code, but hopefully the idea of the function is clear. When I call this function in main, however, I get an error of initializer fails to determine size of ‘mutation’. This is the code I have in main:
int main()
{
char target [] = "Das weisse leid"; //dummy message
char mutation [] = breedMutation(target);
return 0;
}
I need to learn more about pointers and character arrays, which I realise, but I'm trying to learn by example as well.
EDIT: This code, which I'm trying to modify for character arrays, is the basis for breedMutation.
int *f(size_t s){
int *ret=new int[s];
for (size_t a=0;a<s;a++)
ret[a]=a;
return ret;
}
Your error is because you can't declare mutation as a char[] and assign it the value of the char* being returned by breedMutation. If you want to do that, mutation should be declared as a char* and then deleted once you're done with it to avoid memory leaks in a real application.
Your breedMutation function, apart from dynamically allocating an array and returning it, is nothing like f. f simply creates an array of size s and fills each index in the array incrementally starting at 0. breedMutation would just fill the array with 'b' if you didn't have a logic error.
That error is that sizeof(genome); will return the size of a char*, which is generally 4 or 8 bytes on a common machine. You'll need to pass the size in as f does since arrays are demoted to pointers when passed to a function. However, with that snippet I don't see why you'd need to pass a char genome[] at all.
Also, in C++ you're better off using a container such as an std::vector or even std::array as opposed to dynamically allocated arrays (ones where you use new to create them) so that you don't have to worry about freeing them or keeping track of their size. In this case, std::string would be a good idea since it looks like you're trying to work with strings.
If you explain what exactly you're trying to do it might help us tell you how to go about your problem.
The line:
size_t genes = sizeof(genome);
will return the sizeof(char*) and not the number of elements in the genome array. You will need to pass the number of elements to the breedMutation() function:
breedMutation(target, strlen(target));
or find some other way of providing that information to the function.
Hope that helps.
EDIT: assuming it is the number of the elements in genome that you actually want.
Array are very limited.
Prefer to use std::vector (or std::string)
std::string breedMutation(std::string const& genome)
{
std::string mutation;
return mutation;
}
int main()
{
std::string target = "Das weisse leid"; //dummy message
std::string mutation = breedMutation(target);
}
Try replacing the second line of main() with:
char* mutation = breedMutation(target);
Also, don't forget to delete your mutation variable at the end.