I have a List class for char arrays. And I want to push back N arrays from 'a' to 'a..a'.
char* x;
SList list;
for (unsigned int i = 1; i < n+1; i++) {
x = new char[i+1];
for (unsigned int j = 0; j < i; j++) {
x[j] = 'a';
}
x[i] = '\0';
list.push_back(&x);
}
But every time, x has the same address. And in result, my List object contains N pointers to the same address.
Is there a way to push back these arrays in loop with correct memory allocation?
Before asking found this question, but it doesn't provide a cool solution for my problem.
In each iteration of the loop x = new char[i+1]; returns a different address, which is stored in x. So the value of x changes in each iteration, yet, the address of x doesn't.
Did you mean
list.push_back(x);
to add the address of the newly allocated memory? Of course this would require you to change the type of list the a collection of char *.
It must be mentioned that dereferencing &x after x goes out of scope will lead to undefined behaviour, because x doesn't exist anymore. You fill list with dangling pointers.
Finally I'd like to mention that you could avoid the nasty manual memory management and simply use a std::vector<std::string>.
std::vector<std::string> list;
for (unsigned int i = 1; i < n+1; i++) {
std::string newString(i, 'a'); // string with i times 'a'
list.push_back(newString);
}
Ok. I found a pretty straightforward solution:
char** x = new char*[n];
SList sl;
for (unsigned int i = 0; i < n; i++) {
x[i] = new char[i+1];
for (unsigned int j = 0; j < i; j++) {
x[i][j] = 'a';
}
x[i][i] = '\0';
sl.push_back(&x[i]);
}
With having N addresses to store pointers to arrays. I can just add their addresses to my list object
Related
I am trying to create a merge function for two array structures in c++ but am coming up with a bad access error that I don't know how to solve. The error comes up when I am trying to swap the element in the smaller array into the larger, merged array. The code doesn't even go through a single iteration. All three of i, j, and k remain at 0. Any help would be greatly appreciated! Here is the code:
struct Array
{
int *A;
int size;
int length;
};
void display(Array arr){
for (int i = 0; i < arr.length; i++)
std::cout << arr.A[i] << std::endl;
}
Array merge(Array arr1, Array arr2){
Array arr3;
arr3.length = arr1.length + arr2.length;
arr3.size = arr1.length + arr2.length;
int i = 0, j =0, k =0;
while(i <arr1.length && j < arr2.length){
if (arr1.A[i] < arr2.A[j])
{
arr3.A[k] = arr1.A[i]; //(The error is displayed here: Thread 1: EXC_BAD_ACCESS (code=1, address=0x28))
k++;
i++;
}
else if (arr2.A[j] < arr1.A[i])
{
arr3.A[k] = arr2.A[j];
k++;
j++;
}
}
for (; i< arr1.length; i++)
{
arr3.A[k]=arr1.A[i];
k++;
}
for (; i< arr2.length; j++)
{
arr3.A[k]=arr2.A[j];
k++;
}
return arr3;
}
int main() {
Array arr1;
arr1.size = 10;
arr1.length = 5;
arr1.A = new int[arr1.size];
arr1.A[0]= 2;
arr1.A[1]= 6;
arr1.A[2]= 10;
arr1.A[3]= 15;
arr1.A[4]= 25;
Array arr2;
arr2.size = 10;
arr2.length = 5;
arr2.A = new int[arr2.size];
arr2.A[0]= 3;
arr2.A[1]= 4;
arr2.A[2]= 7;
arr2.A[3]= 18;
arr2.A[4]= 20;
Array arr3 = merge(arr1, arr2);
display(arr3);
return 0;
}
Your Array arr3 does not allocate any memory for its int *A field. It's natural that it would not work.
Anyway, your implementation of Array is very poor. Don't reimplement arrays unless you have a good reason; use std::vector instead.
If you really need to implement an Array on your own, then learn about encapsulation, make a class with a constructor, and allocate/delete your data (*A) field properly. Remember, using pointers and heap memory without understanding them is a recipe for disaster.
Easy: arr3.A is not initialized. It's a pointer. What does it point to?
Suggestion: learn about dynamic memory allocation.
I have an array called int **grid that is set up in Amazon::initGrid() and is made to be a [16][16] grid with new. I set every array value to 0 and then set [2][2] to 32. Now when I leave initGrid() and come back in getGrid() it has lost its value and is now 0x0000.
I don't know what to try, the solution seems to be really simple, but I'm just not getting it. Somehow the data isn't being kept in g_amazon but I could post the code.
// Returns a pointer to grid
int** Amazon::getGridVal()
{
char buf[100];
sprintf_s(buf, "Hello %d\n", grid[2][2]);
return grid;
}
int Amazon::initGrid()
{
int** grid = 0;
grid = new int* [16];
for (int i = 0; i < 16; i++)
{
grid[i] = new int[16];
for (int j = 0; j < 16; j++)
{
grid[i][j] = 0;
}
}
grid[2][2] = 32;
return 0;
}
int **grid;
g_amazon = Amazon::getInstance();
g_amazon->initGrid();
grid = g_amazon->getGridVal();
for (int i = 0; i < 16; i++)
{
for (int j = 0; j < 16; j++)
{
int index;
index = (width * 4 * i) + (4 * j);
int gridval;
gridval = grid[i][j];
lpBits[index] = gridval;
lpBits[index + 1] = gridval;
lpBits[index + 2] = gridval;
}
}
It crashes when I run it at the line where sprintf_s prints out [2][2] and it also crashes when I get to gridval = grid[i][j] because it's at memory location 0x000000.
The variable
int** grid
in the initGrid() function is a local variable. Edit** When the function returns the variable is popped off the stack. However, since it was declared with the new operator the memory still exists on the heap; it is simply just not pointed to by your global grid variable.
#Dean said in comment:
I have grid as an int** grid; in class Amazon {}; so shouldn't it stay in memory or do I need a static var.
That is the problem:
local int **grid; on Amazon::initGrid::
is masking
member int **grid; on Amazon::
as the first context has higher priority in name lookup.
So initGrid() allocates memory referenced only by a local pointer. That pointer no longer exists when you return from this function, Amazon::grid was never touched on initialization and you're also left with some bad memory issues.
So, as commented by #Remy-Lebeau, I also suggest
Consider using std::vector> or std::array, 16> instead. There is no good reason to use new[] manually in this situation.
Ok, so I'm quite new to C++ and I'm sure this question is already answered somewhere, and also is quite simple, but I can't seem to find the answer....
I have a custom array class, which I am using just as an exercise to try and get the hang of how things work which is defined as follows:
Header:
class Array {
private:
// Private variables
unsigned int mCapacity;
unsigned int mLength;
void **mData;
public:
// Public constructor/destructor
Array(unsigned int initialCapacity = 10);
// Public methods
void addObject(void *obj);
void removeObject(void *obj);
void *objectAtIndex(unsigned int index);
void *operator[](unsigned int index);
int indexOfObject(void *obj);
unsigned int getSize();
};
}
Implementation:
GG::Array::Array(unsigned int initialCapacity) : mCapacity(initialCapacity) {
// Allocate a buffer that is the required size
mData = new void*[initialCapacity];
// Set the length to 0
mLength = 0;
}
void GG::Array::addObject(void *obj) {
// Check if there is space for the new object on the end of the array
if (mLength == mCapacity) {
// There is not enough space so create a large array
unsigned int newCapacity = mCapacity + 10;
void **newArray = new void*[newCapacity];
mCapacity = newCapacity;
// Copy over the data from the old array
for (unsigned int i = 0; i < mLength; i++) {
newArray[i] = mData[i];
}
// Delete the old array
delete[] mData;
// Set the new array as mData
mData = newArray;
}
// Now insert the object at the end of the array
mData[mLength] = obj;
mLength++;
}
void GG::Array::removeObject(void *obj) {
// Attempt to find the object in the array
int index = this->indexOfObject(obj);
if (index >= 0) {
// Remove the object
mData[index] = nullptr;
// Move any object after it down in the array
for (unsigned int i = index + 1; i < mLength; i++) {
mData[i - 1] = mData[i];
}
// Decrement the length of the array
mLength--;
}
}
void *GG::Array::objectAtIndex(unsigned int index) {
if (index < mLength) return mData[index];
return nullptr;
}
void *GG::Array::operator[](unsigned int index) {
return this->objectAtIndex(index);
}
int GG::Array::indexOfObject(void *obj) {
// Iterate through the array and try to find the object
for (int i = 0; i < mLength; i++) {
if (mData[i] == obj) return i;
}
return -1;
}
unsigned int GG::Array::getSize() {
return mLength;
}
I'm trying to create an array of pointers to integers, a simplified version of this is as follows:
Array array = Array();
for (int i = 0; i < 2; i++) {
int j = i + 1;
array.addObject(&j);
}
Now the problem is that the same pointer is used for j in every iteration. So after the loop:
array[0] == array[1] == array[2];
I'm sure that this is expected behaviour, but it isn't quite what I want to happen, I want an array of different pointers to different ints. If anyone could point me in the right direction here it would be greatly appreciated! :) (I'm clearly misunderstanding how to use pointers!)
P.s. Thanks everyone for your responses. I have accepted the one that solved the problem that I was having!
I'm guessing you mean:
array[i] = &j;
In which case you're storing a pointer to a temporary. On each loop repitition j is allocated in the stack address on the stack, so &j yeilds the same value. Even if you were getting back different addresses your code would cause problems down the line as you're storing a pointer to a temporary.
Also, why use a void* array. If you actually just want 3 unique integers then just do:
std::vector<int> array(3);
It's much more C++'esque and removes all manner of bugs.
First of all this does not allocate an array of pointers to int
void *array = new void*[2];
It allocates an array of pointers to void.
You may not dereference a pointer to void as type void is incomplete type, It has an empty set of values. So this code is invalid
array[i] = *j;
And moreover instead of *j shall be &j Though in this case pointers have invalid values because would point memory that was destroyed because j is a local variable.
The loop is also wrong. Instead of
for (int i = 0; i < 3; i++) {
there should be
for (int i = 0; i < 2; i++) {
What you want is the following
int **array = new int *[2];
for ( int i = 0; i < 2; i++ )
{
int j = i + 1;
array[i] = new int( j );
}
And you can output objects it points to
for ( int i = 0; i < 2; i++ )
{
std::cout << *array[i] << std::endl;
}
To delete the pointers you can use the following code snippet
for ( int i = 0; i < 2; i++ )
{
delete array[i];
}
delete []array;
EDIT: As you changed your original post then I also will append in turn my post.
Instead of
Array array = Array();
for (int i = 0; i < 2; i++) {
int j = i + 1;
array.addObject(&j);
}
there should be
Array array;
for (int i = 0; i < 2; i++) {
int j = i + 1;
array.addObject( new int( j ) );
}
Take into account that either you should define copy/move constructors and assignment operators or define them as deleted.
There are lots of problems with this code.
The declaration void* array = new void*[2] creates an array of 2 pointers-to-pointer-to-void, indexed 0 and 1. You then try to write into elements 0, 1 and 2. This is undefined behaviour
You almost certainly don't want a void pointer to an array of pointer-to-pointer-to-void. If you really want an array of pointer-to-integer, then you want int** array = new int*[2];. Or probably just int *array[2]; unless you really need the array on the heap.
j is the probably in the same place each time through the loop - it will likely be allocated in the same place on the stack - so &j is the same address each time. In any case, j will go out of scope when the loop's finished, and the address(es) will be invalid.
What are you actually trying to do? There may well be a better way.
if you simply do
int *array[10];
your array variable can decay to a pointer to the first element of the list, you can reference the i-th integer pointer just by doing:
int *myPtr = *(array + i);
which is in fact just another way to write the more common form:
int *myPtr = array[i];
void* is not the same as int*. void* represent a void pointer which is a pointer to a specific memory area without any additional interpretation or assuption about the data you are referencing to
There are some problems:
1) void *array = new void*[2]; is wrong because you want an array of pointers: void *array[2];
2)for (int i = 0; i < 3; i++) { : is wrong because your array is from 0 to 1;
3)int j = i + 1; array[i] = *j; j is an automatic variable, and the content is destroyed at each iteration. This is why you got always the same address. And also, to take the address of a variable you need to use &
I have an array of chars. n is array's length
char tab[n];
cin.get(tab, n);
cout<<tab<<"\0"<<endl;
then I create second array
char* t = new char[n];
for(int i = 0; tab[i] != '\0'; i++){
t[i] = tab[i];
}
I would like to have pointers to corresponding element. I mean t[2] contains addres of tab[2]. Then I would like to sort array t so tab stays as it was and only t shuffles but when I make change in the array t for example
t[2] = 'a';
I loose t[3]t[4]....
EDIT:
do{
for(int i = 0; i < n -1; i++){
if(t[i] > t[i+1]){
char* x = &t[i];
t[i] = t[i+1];
t[i+1] = *x
}
n--;
}
}while(n>1);
I would like to have pointers to corresponding element. I mean t2 contains addres of tab2.
So try this:
char** t = new (char*)[n];
for(int i = 0; tab[i] != '\0'; i++){
t[i] = tab+i; // or `&(tab[i])`
To sort the t array based on tab values, use *t[i] to access character values stored in location i of the t array.
See here and here for good tutorials about C pointers.
CASE1:
int nrows=5;
int ncols=10;
int **rowptr;
rowptr=new int*;
for(int rows=0;rows<nrows;rows++) {
for(int cols=0;cols<ncols;cols++) {
*rowptr=new int;
}
}
CASE2:
int nrows=5;
int ncols=10;
int **rowptr;
for(int rows=0;rows<nrows;rows++) {
rowptr=new int*;
for(int cols=0;cols<ncols;cols++) {
*rowptr=new int;
}
}
I am able to insert and print values using both ways. What is the difference in initializations?
What is the difference?
#1 just allocates memory enough to hold a integer pointer and not an array of integer pointers.
#2 Causes a memory leak by just overwritting the memory allocation of the previous iteration.
I am able to insert and print values using both the ways
Memory leaks and Undefined behaviors may not produce immediate observale erroneous results in your program but they sure are good cases of the Murphy's Law.
The correct way to do this is:
int nrows = 5;
int ncols = 10;
//Allocate enough memory for an array of integer pointers
int **rowptr = new int*[nrows];
//loop through the array and create the second dimension
for (int i = 0;i < nrows;i++)
rowptr[i] = new int[ncols];
You have a memory leak in both cases.
The proper way to initialize such a "2d" array is
int** arr = new int*[nrows];
for (int i = 0; i < nrows; i++)
arr[i] = new int[ncols];
Note however, that it isn't a 2d array as defined by C/C++. It may not, and probably will not, be consecutive in memory. Also, the assembly code for accessing members is different.
In your case, the accessing by indexing is equivalent to *(*(arr+i)+j)
And in the case of a 2d array it's *(arr + N_COLS*i + j) when N_COLS is a compile time constant.
If you want a true 2d array you should do something like this:
int (*arr)[N_COLS] = (int(*)[N_COLS])(new int[N_ROWS * N_COLS])
You'd better use 1d array to manage 2d array
int **x = new int*[nrows];
x[0] = new int[nrows*ncols];
for (int i = 1; i < nrows; i++)
x[i] = x[i-1] + ncols;
for (int i = 0; i < nrows; i++)
for (int j = 0; j < ncols; j++)
x[i][j] = 0;
delete [] x[0];
delete [] x;