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.
Related
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.
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 ;)
I have 2 doubts regarding basics of pointers usage.
With the following code
int (*p_b)[10];
p_b = new int[3][10];
// ..do my stuff
delete [] p_b
p_b is pointing to an array of 3 elements, each having fixed-size length of 10 int.
Q1:
How to declare p_b if I want that each element be a pointer to a fixed array size?
Basically I want the following
p_b[0] = pointer to a fixed-array size of 10
p_b[1] = pointer to a fixed-array size of 10
// ... and so on
I was thinking to int (** p_b)[10] but then I don't know how to use new to allocate it? I would like to avoid falling back to more general int** p_b
Q2:
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10] ? To free memory I have to call delete[] while I cannot find an expression where I can call only simply delete.
p_b is pointing to an array of 3 elements, each having fixed-size length of 10 int.
How to declare p_b if I want that each element be a pointer to a fixed array size?
Does your first sentence not completely cover that question?
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10]? To free memory I have to call delete[] while I cannot find an expression where I can call only simply delete.
I completely do not understand why this is a problem, but you could do it by wrapping your array inside another type... say std::array, boost::array or std::vector.
First of all, if your new expression has square brackets (new somtype[somesize]), your delete has to have square brackets as well (delete [] your_pointer).
Second, right now you've defined p_b to be a single pointer to some data. If what you really want is an array of pointers, then you need to define it as an array. Since you apparently want three independent arrays, you'll have to allocate each of them separately. It's probably easiest if you start with a typedef:
typedef int *p_int;
p_int p_b[3];
Then you'll allocate your three arrays:
for (int i=0; i<3; i++)
p_b[i] = new int[10];
To delete those, you'll need to delete each one separately:
for (int i=0; i<3; i++)
delete [] p_b[i];
I definitely agree with #Tomalak that you should almost never mess with things like this yourself though. It's not clear what you really want to accomplish, but it's still pretty easy to guess that chances are quite good that a standard container is likely to be a simpler, cleaner way to do it anyway.
Here's an example of how to implement Q1:
int main()
{
typedef int foo[10];
foo* f = new foo[3];
f[0][5] = 5;
f[2][7] = 10;
delete [] f;
}
As for Q2, the only way to delete memory allocated with new[] is with delete[]. If you personally don't want to write delete [], you can use a vector or another STL container. Really, unless this is some hardcore uber-optimisation, you should be using vectors anyway. Never manage memory manually unless you are absolutely forced to.
To use a raw pointer to manage a 2-d array you must first create a pointer to a pointer to array element type that will point to each row of the array. Next, each row pointer must be assigned to the actual array elements for that row.
int main()
{
int **p;
// declare an array of 3 pointers
p = new int*[3];
// declare an array of 10 ints pointed to by each pointer
for( int i = 0; i < 3; ++i ) {
p[i] = new int[10];
}
// use array as p[i][j]
// delete each array of ints
for( int i = 0; i < 3; ++i ) {
delete[] p[i];
}
// delete array of pointers
delete[] p;
}
A far easier solution is to use std::array. If your compiler does not provide that class you can use std::vector also.
std::array<std::array<int,10>,3> myArr;
myArr[0][0] = 1;
For Q1, I think you want
int (*p[3])[10];
Try cdecl when you're unsure.
Your other question seems to be well answered by other answers.
regards,
Yati Sagade
Actually, nobody posted an answer to your exact question, yet.
Instead of
int (*p_arr)[10] = new int[3][10];
// use, then don't forget to delete[]
delete[] p_arr;
I suggest using
std::vector<std::array<int, 10>> vec_of_arr(3);
or if you don't need to move it around and don't need runtime length:
std::array<std::array<int, 10>, 3> arr_of_arr;
Q1
How to declare p_b if I want that each element be a pointer to a fixed array size?
int(**pp_arr)[10] = new std::add_pointer_t<int[10]>[3];
for (int i = 0; i < 3; ++i)
pp_arr[i] = new int[1][10];
// use, then don't forget to delete[]
for (int i = 0; i < 3; ++i)
delete[] pp_arr[i];
delete[] pp_arr;
The modern variant of that code is
std::vector<std::unique_ptr<std::array<int, 10>>> vec_of_p_arr(3);
for (auto& p_arr : vec_of_p_arr)
p_arr = std::make_unique<std::array<int, 10>>();
or if you don't need to move it around and don't need runtime length:
std::array<std::unique_ptr<std::array<int, 10>>, 3> arr_of_p_arr;
for (auto& p_arr : arr_of_p_arr)
p_arr = std::make_unique<std::array<int, 10>>();
Q2
Is per my original code sample above, how to call new so that p_b points to a unique fixed-size array of 10 int other than calling p_b = new int[1][10]?
Not without wrapping the array into another type.
std::array<int, 10>* p_arr = new std::array<int, 10>;
// use, then don't forget to delete
delete p_arr;
You can replace std::array<int, 10> with your favourite array-wrapping type, but you cannot replace it with a fixed-size array alias. The modern variant of that code is:
auto p_arr = std::make_unique<std::array<int, 10>>();
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++?