int (**test)[4][4] = new ???[64];
for (int i = 0; i < 32; ++i)
{
test[i] = new int[4][4][4];
}
I'm trying to create a "list" of pointers that will be initialized to NULL and then later assigned the address of a new multidimensional array of int. The for loop will (eventually) vary in number of iterations, anywhere from 0 to the full 64. I expect to end up with an array of pointers where some are valid and the rest are NULL. The problem is that I can't figure out the syntax for allocating this array of pointers. Basically, what could I put in place of those question marks?
In the name of readability, may I suggest using a typedef?
typedef int (*t)[4][4];
t* test = new t[64];
You will thank me next week when you have to maintain that horrible piece of code ;)
Related
I have a dynamically populated array of strings in C++:
string** A;
it is populated like this:
A = new string*[size1];
and then:
for (unsigned int i = 0; i < size1; i++)
{
A[i] = new string[size2];
for (unsigned int j = 0; j < size2; j++)
{
A[i][j] = whatever[j];
}
}
elsewhere, I want to find out the dimensions (size1 and size2).
I tries using this:
sizeof(A[i]) / sizeof(A[i][0])
but it doesn't work.
Any ideas ?
Thanks
When you allocate memory via new T[N], the value N is not stored anywhere . If you need to know it later, you will need to keep track of it in your code.
There are pre-existing classes for allocating memory that also remember the length that was allocated. In your code:
vector<vector<string>> A(size1, vector<string>(size2));
// (code to populate...)
then you can access A.size() to get size1, and A[0].size() to get size2.
If the dimensions are known at compile-time you may use array instead of vector.
It is very simple to find the size of a two dimensional (more exactly of one-dimensional dynamically allocated arrays) array. Just declare it like
std::vector<std::vector<std::string>> A;
and use
std::cout << A.size() << std::endl;
As for your approach then you have to store the sizes in some variables when the array is allocated.
If you are learning C++, I would recommend that you learn Classes. With a class you can encapsulate int variables along with your 2D array that you can use to store the dimensions of your array. For example:
class 2Darray{
string **array;
int rows;
int cols;
}
You can then get the dimensions of your 2Darray object anytime by reading these member variables.
vectors will do this for you behind the scenes but its good for you to learn how to do this.
You can't create an array just using pointer operator. Every array is basically a pointer with allocated memory. That's why compiler wants constant before creating array.
Basically; sizeof(A[i]) won't give you the size of array. Because sizeof() function will return the a pointers size which is points to A[i] location. sizeof(A[i]) / sizeof(A[i][1]) will probably give you 1 because you are basically doing sizeof(int)/sizeof(int*)
So you need to store the boundary yourself or use vectors. I would prefer vectors.
Can't get array dimensions through pointer(s)
I am trying to initialize a multidimensional array in batches and can't seem to make it work or find an example!
The dimensions I am working with are big enough that I don't want to specify them by hand!
More precisely :
int test[5][192];
for(int i = 0; i < 5; i++){
int temp[192] = {...};
test[i] = temp;
}
// use variable test here..
I want to use this method because the temp array is dynamicaly defined depending on variable i.
Is this type of initialization possible?
Should the temp array be in dynamic memory?
Since after the initialization I pass a reference to the first element of test to another function and I am not in control of how the other function passes over the elements I need to keep the data type of an array!
If you want to copy the values of temp array , instead of "=", you should use memory copy
memcpy( test[i], temp, sizeof(temp[192]));
Arrays do not have the copy assignment operator. So this is impossible with arrays.
If you will dynamically allocate each row then in any case you have to store somewhere the number of their elements. So even dynamically allocated arrays are not suitable in this case when the numbers of elements in each row can differ.
You should use standard container std::vector<std::vector<int>> instead.
You dont use the second dimension in the array for the test array. you just write test[], but you must write test[][].
I thing you don't must use the temp array. You can initialize you array direct, without using a temp array.
I'm not absolutly sure, but memcpy are just using for one diemnsionalarrays ant not for multidimensional arrays
if you know, Temp and test[i] are two pointer which point to memory
so if you print temp or test[i] you will see the address of where they start on memory.
in your code you lose the address of test[i] because you changed pointer test[i] to temp
and now both of them are pointing to the same place where temp begin on there!
I am a c++ newbie. While learning I came across this.
if I have a pointer like this
int (*a)[2][3]
cdecl.org describe this as declare a as pointer to array 2 of array 3 of int:
When I try
int x[2][3];
a = &x;
this works.
My question is how I can initialize a when using with new() say something like
a = new int [] [];
I tried some combinations but doesn't get it quite right.
Any help will be appreciated.
You will have to do it in two steps - first allocate an array of pointers to pointers(dynamically allocated arrays) and then, allocate each of them in turn. Overall I believe a better option is simply to use std::vector - that is the preferred C++ way of doing this kind of things.
Still here is an example on how to achieve what you want:
int a**;
a = new int*[2];
for (int i =0; i< 2;++i){
a[i] = new int[3]
}
... use them ...
// Don't forget to free the memory!
for (int i = 0; i< 2; ++i) {
delete [] a[i];
}
delete [] a;
EDIT: and as requested by Default - the vector version:
std::vector<std::vector<int> > a(2, std::vector<int>(3,0));
// Use a and C++ will take care to free the memory.
It's probably not the answer you're looking for, but what you
need is a new expression whose return type is (*)[2][3] This
is fairly simple to do; that's the return type of new int
[n][2][3], for example. Do this, and a will point to the
first element of an array of [2] of array of [3] int. A three
dimensional array, in sum.
The problem is that new doesn't return a pointer to the top
level array type; it returns a pointer to the first element of
the array. So if you do new int[2][3], the expression
allocates an array of 2 array of 3 int, but it returns
a pointer to an array of 3 int (int (*a)[3]), because in C++,
arrays are broken (for reasons of C compatibility). And there's
no way of forcing it to do otherwise. So if you want it to
return a pointer to a two dimensional array, you have to
allocate a three dimensional array. (The first dimension can be
1, so new [1][2][3] would do the trick, and effectively only
allocate a single [2][3].)
A better solution might be to wrap the array in a struct:
struct Array
{
int data[2][3];
};
You can then use new Array, and everything works as expected.
Except that the syntax needed to access the array will be
different.
First off: I know that new is the C++ way of doing this. I am simply showing that there is more than one way to reproduce this error, and both are incredibly frustrating.
I have two forms of this source file. I'm trying to debug yet another programming assignment, but I'm not asking for help on that. Basically, I'm trying to re-implement set as a class with fields for size and a pointer to an int array. Here's the code using new:
testnew.cpp
int main()
{
int size = 1;
int *elements = new int[size];
elements[0] = 0;
size++;
int * temp = new int[size];
for (int i = 0; i < (size - 1); i++)
{
temp[i] = elements[i];
}
delete[] elements;
temp[size] = size;
elements = temp;
elements[1] = 1;
delete[] elements;
}
and again, using the less-preferable alloc functions:
testalloc.cpp
int main()
{
int size = 1;
int * elements = (int *) malloc(sizeof(int) * size);
elements[0] = 0;
size++;
elements =(int *) realloc(elements,size * sizeof(int));
elements[1] = 1;
free(elements);
}
In both cases, my goal is to create an array, and then append to it. However, in both cases, after building and running it in Visual Studio 2010, the array is not grown at all, and only has 1 "slot" for my items to go into. In VS's debugger, I have watches on the elements array pointer; attached is a screenshot. It is the same for both versions of the code.
-- the breakpoint is at the delete[]/free() call.
Seriously, what am I doing wrong? This has to be a logic error, and I've combed through a quarter dozen examples of both malloc/realloc and new, and read and re-read my textbook, and I can't see what's wrong!
I feel like I should have a line that splits the allocated memory into an array, but doesn't the new int[] call do that?
Besides any other issues with the code, you must either do what JoeG mentions and have multiple watches, or set a watch like this:
elements,5
will result in:
I don't know if there is an updated list for VS2010, but this still works:
Symbols for Watch Variables
And:
View array in Visual Studio debugger?
Other answers have pointed out the bug in your code (temp[size] = size;), however your confusion is coming from the fact that you're misreading the debugger's output.
As far as the type system and debugger are concerned, elements isn't an array, it's a pointer. The debugger has no way of knowing whether or not it's a pointer to the first element in an array or a pointer to a single element.
If you want to see the value of elements[x] in your debugger, use the expression *(elements+x).
In the first example, this is wrong:
temp[size] = size;
This will translate to:
temp[2] = 2;
And since arrays are zero-indexed you are writing outside your allocated area.
First of all C++ is 0 indexed, so temp[size] = size; is an off by one error. And to answer your question, the type of elements is int*. That it's actually an array is not knowledge which is available to VS without code analysis. So, what you need to do is either use for example std::vector<>, boost::array or make sure that your array never detoriates to int*.
The file testnew.cpp allocates memory for size elements in temp but then sets temp[size], which is the size+1 element. Perhaps this should be
temp[i] = size;
The file testalloc.cpp commits the common error of reassigning memory to the same variable without verifying that thte call to realloc succeeds:
elements =(int *) realloc(elements,size * sizeof(int));
If realloc fails, elements will be set to null, and its original memory will be orphaned.
int *x = new int[5]();
With the above mentality, how should the code be written for a 2-dimensional array - int[][]?
int **x = new int[5][5] () //cannot convert from 'int (*)[5]' to 'int **'
In the first statement I can use:
x[0]= 1;
But the second is more complex and I could not figure it out.
Should I use something like:
x[0][1] = 1;
Or, calculate the real position then get the value
for the fourth row and column 1
x[4*5+1] = 1;
I prefer doing it this way:
int *i = new int[5*5];
and then I just index the array by 5 * row + col.
You can do the initializations separately:
int **x = new int*[5];
for(unsigned int i = 0; i < 5; i++)
x[i] = new int[5];
There is no new[][] operator in C++. You will first have to allocate an array of pointers to int:
int **x = new int*[5];
Then iterate over that array. For each element, allocate an array of ints:
for (std::size_t i = 0; i < 5; ++i)
x[i] = new int[5];
Of course, this means you will have to do the inverse when deallocating: delete[] each element, then delete[] the larger array as a whole.
This is how you do it:
int (*x)[5] = new int[7][5] ;
I made the two dimensions different so that you can see which one you have to use on the lhs.
Ff the array has predefined size you can write simply:
int x[5][5];
It compiles
If not why not to use a vector?
There are several ways to accomplish this:
Using gcc's support for flat multidimensional arrays (TonyK's answer, the most relevant to the question IMO). Note that you must preserve the bounds in the array's type everywhere you use it (e.g. all the array sizes, except possibly the first one), and that includes functions that you call, because the produced code will assume a single array. The allocation of $ new int [7][5] $ causes a single array to be allocated in memory. indexed by the compiler (you can easily write a little program and print the addresses of the slots to convince yourself).
Using arrays of pointers to arrays. The problem with that approach is having to allocate all the inner arrays manually (in loops).
Some people will suggest using std::vector's of std::vectors, but this is inefficient, due to the memory allocation and copying that has to occur when the vectors resize.
Boost has a more efficient version of vectors of vectors in its multi_array lib.
In any case, this question is better answered here:
How do I use arrays in C++?