I am currently reading some C++ source code, and I came across this:
double **out;
// ... lots of code here
// allocate memory for out
out = new double*[num];
Not entirely sure what it does, or what it means. Is it a pointer... to another pointer?
There is also the following:
double ***weight;
// allocate memory for weight
weight = new double**[numl];
I am quite confused :P, any help is appreciated.
new double*[num] is an array of double pointers i.e. each element of the array is a double*. You can allocate memory for each element using out[i] = new double; Similarly weight is an array of double**. You can allocate the memory for each weight element using new double*[num] (if it is supposed to be an array of double*)
It's a pointer to pointer to double. Or array of pointers to double. Or if every pointer itself allocates array it might be a matrix.
out = new double*[num]; // array of pointers
Now it depents if out[0] is allocated like this:
out[0] = new double; // one double
or like this:
out[0] = new double[num]; // now you've got a matrix
Actually, writing
double*[] out;
is in C/C++ equal to
double** out;
and it means an array of pointers to double. Or a pointer to pointers of double. Because an array is nothing more than just a pointer. So this is in essence a two-dimensional array.
You could as well write
double[][] out;
And likewise, adding another pointer level, will add another dimension to your array.
So
double ***weight;
is actually a pointer to a three-dimensional array.
Basically both of your code fragments allocate array of pointers. For allocation it does not matters to what. Correct declaration is needed only for type checks. Square bracjets should be read separately and means only it is array.
Consider following code as quick example:
#include <stdio.h>
int main()
{
unsigned num = 10;
double **p1, ***p2;
p1 = new double*[num];
p2 = new double**[num];
printf("%d\n", sizeof(p1));
printf("%d\n", sizeof(p2));
delete [] p1;
delete [] p2;
return 0;
}
Yes, both are just pointers. And memory allocated is sizeof(double*) * num.
Related
I am confused about this line in a C++ program. The idea of the program is to check whether a 4x4 array is symmetric or not. This part of the code declares a 2D array, which I do not understand.
int** array = new int*[n];
Although, there is another question similar to this but it is about single pointer which I get.
int *array = new int[n];
I do not understand the double pointer. Kindly explain.
How do you create a single pointer array? You do this:
int* myArray = new int[n];
What does this mean? It has two parts. First part is reserving a pointer int* we call it myArray, and the second part is that you reserve n elements, each with size int in memory (this is an array, right?), and you take the address of that array and you save it in the variable myArray.
Now you want a 2D array, which is an array of an array. So Every element of this new array of array is one of these, that we talked about up there. How do we reserve this? We do:
new int*[n];
Because we are reserving n slots, each with type int*, that we talked about before.
Now what is the type of the return value? It's an array of an array, or a "pointer to an array, and the latter is also a pointer to an array", so you write it as
(int*)*
Or
int**
so it becomes
int** array = new int*[n];
int** array is a pointer to a pointer to an int. So by doing this:
int** array = new int*[n];
you are creating a section of memory that holds n int* pointers and pointing array at that memory. For each of these pointers that you have created, it is possible to create a set of ints like so:
for (auto i = 0; i < n; ++i)
array[i] = new int[n];
The resulting memory will look like this:
array -> [ int* | int * | .... n
[int | int | ...][int | int | ...][ ... n
This is however, much much easier if you use some of the std things in c++, ie a std::vector :
std::vector<std::vector<int>> arr(std::vector<int>(0, n), n);
and you are done ...
I need to init/use a double ** (decleared in my header):
double **pSamples;
allocating (during the time) a matrix of NxM, where N and M are get from two function:
const unsigned int N = myObect.GetN();
const unsigned int M = myObect.GetM();
For what I learnt from heap and dynamic allocation, I need keyword new, or use STL vector, which will manage automatically allocate/free within the heap.
So I tried with this code:
vector<double> samplesContainer(M);
*pSamples[N] = { samplesContainer.data() };
but it still says I need a constant value? How would you allocate/manage (during the time) this matrix?
The old fashioned way of initializing a pointer to a pointer, is correctly enough with the new operator, you would first initialize the the first array which is a pointer to doubles (double*), then you would iterate through that allocating the next pointer to doubles (double*).
double** pSamples = new double*[N];
for (int i = 0; i < N; ++i) {
pSambles[i] = new double[M];
}
The first new allocates an array of double pointers, each pointer is then assigned to the array of pointers allocated by the second new.
That is the old way of doing it, remember to release the memory again at some point using the delete [] operator. However C++ provide a lot better management of sequential memory, such as a vector which you can use as either a vector of vectors, or simply a single vector capable of holding the entire buffer.
If you go the vector of vector way, then you have a declaration like this:
vector<vector<double>> samples;
And you will be able to reference the elements using the .at function as such: samples.at(2).at(0) or using the array operator: samples[2][0].
Alternatively you could create a single vector with enough storage to hold the multidimensional array by simply sizing it to be N * M elements large. However this method is difficult to resize, and honestly you could have done that with new as well: new double[N * M], however this would give you a double* and not a double**.
Use RAII for resource management:
std::vector<std::vector<double>> samplesContainer(M, std::vector<double>(N));
then for compatibility
std::vector<double*> ptrs(M);
for (std::size_t i = 0; i != M; ++i) {
ptrs[i] = samplesContainer[i].data();
}
And so pass ptrs.data() for double**.
samplesContainer.data() returns double*, bur expression *pSamples[N] is of type double, not double*. I think you wanted pSamples[N].
pSamples[N] = samplesContainer.data();
I need help understanding the syntax of multidimensional arrays in C++. In the book I'm learning C++ from, the code snippet looks like this:
typedef int* IntArrayPtr;
IntArrayPtr *m = new IntArrayPtr[num_rows];
for(int i = 0; i < rows; i++){
m[i] = new int[num_columns]
}
My question is this: Why is there a star infront of the m? To me when I see
new IntArrayPtr[num_rows];
that's enough information to tell the compiler that it's an array of pointers that point to int. The star just makes it confusing. Is there something I'm missing here?
Keep in mind that what you have when you do new IntArrayPtr[num_rows] is an array of IntArrayPtrs. In C new[], "allocates size bytes of storage, suitably aligned to represent any object of that size, and returns a non-null pointer to the first byte of this block." So new[] is returning you a pointer to the first element of your array.
For example if num_rows is 3 this is what gets allocated in memory:
m --> [IntArrayPtr]
[IntArrayPtr]
[IntArrayPtr]
m being a pointer is what allows you to use the index operator on it: m[1] returns you the second IntArrayPtr in m.
m is the pointer to pointer which handles starting points of columns as a row. Example of this example in c
int **arr=(int**)malloc(num_rows * (sizeof(int*) ) );
for(i=0;i<row;i++)
{
arr[i]=(int*)malloc(sizeof(int)*col_rows);
}
if we use typedef,
typedef int* IntArrayPtr;
IntArrayPtr *arr=(IntArrayPtr*)malloc(num_rows * (sizeof(IntArrayPtr) ) );
for(i=0;i<row;i++)
{
arr[i]=(IntArrayPtr)malloc(sizeof(int)*col_rows);
}
As you see, firstly we create a pointer array to hold pointers of columns. After that we allocate a place for every columns of array and assign starting points of one dimensional column arrays to one dimensional pointer array.
The problem that I have is to create a specific matrix.
I have to use an array called for example ptr with x pointers. Each pointer in this array should point to a new array (in this case, an int array; each array is a new line in the matrix then).
All x arrays should be created with new; in the end, it should be possible to access the matrix with ptr[a][b] easily.
After a lot of trying and failing, I hope that someone can help me out.
Thank you in advance!
Since this is obviously homework, let me give you a better answer for your sake to go alongside the accepted one.
std::vector<std::vector<int>> matrix(10, std::vector<int>(10));
// ^ ^ ^
// Column count ______| |________________|
// |
// |___ Each column is
// initialized with
// a vector of size 10.
That's a 10x10 matrix. Since we're using vectors, the sizes are dynamic. For statically sized arrays, you can use std::array if you want. Also, here's the reference for std::vector.
If the number of pointers in the array is known, you could simply use a raw array of pointers to int:
int* my_array[10]; // 10 int*
Then you should allocate memory individually for each pointer in the array using usually a for loop:
for(int i=0; i<10; i++){
// each int* in the array will point to an area equivalent to 10 * sizeof(int)
my_array[i] = new int[10];
}
On the other hand, if you don't know the size of the array, then you need a pointer to pointers:
int** ptr_to_ptr = new int*[10];
Note that I am allocating space for 10 int* and not int.
Remember to deallocate the memory allocated above also for the internal pointers, once you don't need that memory anymore.
Here is my simple code
arrayfunc() should store some numbers in an array, and return the pointer of this array
to main function where the content of the array would be printed
What is the problem with my code?
It only returns the pointer to the first element of the array
Any help will be appreciated.
Thanks in advance.
#include <iostream>
using namespace std;
//The definition of the function should remain the same
int* arrayfunc()
{
int *array[10];
array[0] =new int;
array[1] =new int;
array[2] =new int;
array[3] =new int;
*array[0]=10;
*array[1]=11;
*array[2]=12;
*array[3]=13;
return *array;
}
int main()
{
for(int i=0;i<4;i++)
cout<<*(arrayfunc()+i)<<endl;
return 0;
}
(1) You should allocate your array with new if you want to return it: int* array = new int[10]; [assuming here you want array of ints and not array of int*'s]
(2) to return the pointer to the first element in the array, use return array and not return *array
(3) your array is array of pointers, and not array of ints.
Your array is allocated on stack, so as soon as the function returns, it's freed. So you want to return a pointer to a dead memory.
But you are not doing that, you are just returning the valid (copy of) value of the 0th array item.
So, what you have to do:
The best idea would be to switch to stl containers. You should be using std::vector or something like that.
If you stick to the idea of manual memory management, you have to allocate the array on heap, return it from the function, and perhaps deallocate it in the caller.
Edit:
basically you want the following:
using namespace std;
vector<int> arrayfunc()
{
vector<int> v;
v.push_back(10);
...
return v;
}
...
vector<int> result = arrayfunc();
cout << result[0] << ...
This would be the right C++ way.
(Nitpicking:) You don't need to care about copying the vector, because of the RVO used by all modern C++ compilers.
Allocating an array on heap should be simple, too:
int* array = new int[4];
array[0] = 10;
...
return array;
...
int* array = arrayfunc();
...
delete[] array;
But I would strongly advise to take the former approach (with vector).
This codes seems wrong to me in several levels.
Never return an internal variable of a function. The variable array is only defined in the function, so it should never be returned outside.
Why do you allocate each int by itself with new? I would allocate the entire array at once. If you know the array length and it's constant, consider having it defined statically.
http://msdn.microsoft.com/en-us/library/s1sb61xd.aspx
Just try return array; instead of return *array;