I am creating a 2D-Array in C++ and need to pass this array as a parameter in a function. In my function, I need to access an element from the array in order to save it as a value, i.e.:
int lowestPoint(int **arr, int x, int y, int n) {
minVal = *(*(arr+x)+y); // here is where I'm getting the exception
return minVal;
}
I've tried setting minVal to arr[X][Y] and have tried to pass the array in as other variations instead of just **arr but nothing seems to be working.
The array is initialized in my main function as int arr[x][y] and I pass it into another function by casting it as otherFunc(reinterpret_cast<int **>((*arr)[n]), n), and then from that function, send it to lowestPoint by calling int val = lowestPoint(arr,i,j,n). I think these calls could be problematic but I'm uncertain how to fix it - I really have no experience with 2D arrays in C++ and it's soo much simpler in Java. I keep getting an EXC_BAD_ACCESS error for the array, so if anyone has any idea how to fix that, I'd really appreciate it. Thanks!
EDIT:
"n" is the size of the array; for example if it's a 3x3 array, n = 3. I just initialized the array as int arr[n][n] and then stored elements. I know the actual array itself represents the correct value, it just can't access it once I send it to another function.
When you pass the array to the first function using reinterpret_cast((*arr)[n]), instead of passing the pointer to the actual array, you are passing the value in location [0][n] (by using (*arr)[n]) and casting it to **arr. So in essence you get a new array that points to a random location in memory that is equal to the content of that array slot.
I am not sure what you intended to do, but if you wanted to pass the actual array, just pass arr. If you planned to pass a sub-array, this method is incorrect altogether as you pass an offset inside an array and you will get skewed data.
Hope This helps,
Lior
Related
If i have array of pointer to an array of pointer to int
int **arrs = new int *[n];
and assign each element to new array of pointer to int
for i<n
arrs[i] = new int[size]
Now, when i try to get size of these arrays, give me wrong value
int size = sizeof(arrs[0])/sizeof(int);
It gives me wrong value.
So how can i get the right value ?
You can't. arrs[0] is a pointer, so sizeof(arrs[0]) gives the size of that pointer.
It comes as a surprise to some that there is no way to get the size of an array from only a pointer to that array. You have to store the size somewhere else.
In C++ the simplest and best solution is to use std::vector. Why reinvent the wheel? std::vector has everything you would want from a dynamic array, and should be the first choice in this situation.
for a project using Tensorflow's C API I have to pass a void pointer (void*) to a method of Tensorflow. In the examples the void* points to a 2d array, which also worked for me. However now I have array dimensions which do not allow me to use the stack, which is why I have to use a dynamic array or a vector.
I managed to create a dynamic array with the same entries like this:
float** normalizedInputs;//
normalizedInputs = new float* [noCellsPatches];
for(int i = 0; i < noCellsPatches; ++i)
{
normalizedInputs[i] = new float[no_input_sizes];
}
for(int i=0;i<noCellsPatches;i++)
{
for(int j=0;j<no_input_sizes;j++)
{
normalizedInputs[i][j]=inVals.at(no_input_sizes*i+j);
////
////
//normalizedInputs[i][j]=(inVals.at(no_input_sizes*i+j)-inputMeanValues.at(j))/inputVarValues.at(j);
}
}
The function call needing the void* looks like this:
TF_Tensor* input_value = TF_NewTensor(TF_FLOAT,in_dims_arr,2,normalizedInputs,num_bytes_in,&Deallocator, 0);
In argument 4 you see the "normalizedInputs" array. When I run my program now, the calculated results are totally wrong. When I go back to the static array they are right again. What do I have to change?
Greets and thanks in advance!
Edit: I also noted that the TF_Tensor* input_value holds totally different values for both cases (for dynamic it has many 0 and nan entries). Is there a way to solve this by using a std::vector<std::vector<float>>?
Respectively: is there any valid way pass a consecutive dynamic 2d data structure to a function as void*?
In argument 4 you see the "normalizedInputs" array. When I run my program now, the calculated results are totally wrong.
The reason this doesn't work is because you are passing the pointers array as data. In this case you would have to use normalizedInputs[0] or the equivalent more explicit expression &normalizedInputs[0][0]. However there is another bigger problem with this code.
Since you are using new inside a loop you won't have contiguous data which TF_NewTensor expects. There are several solutions to this.
If you really need a 2d-array you can get away with two allocations. One for the pointers and one for the data. Then set the pointers into the data array appropriately.
float **normalizedInputs = new float* [noCellsPatches]; // allocate pointers
normalizedInputs[0] = new float [noCellsPatches*no_input_sizes]; // allocate data
// set pointers
for (int i = 1; i < noCellsPatches; ++i) {
normalizedInputs[i] = &normalizedInputs[i-1][no_input_sizes];
}
Then you can use normalizedInputs[i][j] as normal in C++ and the normalizedInputs[0] or &normalizedInputs[0][0] expression for your TF_NewTensor call.
Here is a mechanically simpler solution, just use a flat 1d array.
float * normalizedInputs = new float [noCellsPatches*no_input_sizes];
You access the i,j-th element by normalizedInputs[i*no_input_sizes+j] and you can use it directly in the TF_NewTensor call without worrying about any addresses.
C++ standard does its best to prevent programmers to use raw arrays, specifically multi-dimensional ones.
From your comment, your statically declared array is declared as:
float normalizedInputs[noCellsPatches][no_input_sizes];
If noCellsPatches and no_input_sizes are both compile time constants you have a correct program declaring a true 2D array. If they are not constants, you are declaring a 2D Variable Length Array... which does not exist in C++ standard. Fortunately, gcc allow it as an extension, but not MSVC nor clang.
If you want to declare a dynamic 2D array with non constant rows and columns, and use gcc, you can do that:
int (*arr0)[cols] = (int (*) [cols]) new int [rows*cols];
(the naive int (*arr0)[cols] = new int [rows][cols]; was rejected by my gcc 5.4.0)
It is definitely not correct C++ but is accepted by gcc and does what is expected.
The trick is that we all know that the size of an array of size n in n times the size of one element. A 2D array of rows rows of columnscolumns if then rows times the size of one row, which is columns when measured in underlying elements (here int). So we ask gcc to allocate a 1D array of the size of the 2D array and take enough liberalities with the strict aliasing rule to process it as the 2D array we wanted. As previously said, it violates the strict aliasing rule and use VLA in C++, but gcc accepts it.
can i pass a multi-dimension array without passing its size using pointers?
#include <iostream>
using namespace std;
void PrintArray( int(*pInt)[3] )
{
// how to iterate here printing pInt content?
int nRow = sizeof(*pInt) / sizeof(int);
cout << nRow << endl; // the result: 3 not 2! i get the nColumns instead!!?
}
int main()
{
int array[2][3] = { {1,2,3},
{4,5,6}};
int(*pArray)[3] = array;
PrintArray(pArray);
return 0;
}
can anyone add the correct definition of PrintArray to print its elements
is there a way to pass arrays as a pointer and without passing the size?
thank you for throwing a glance
can anyone add the correct definition of PrintArray to print its elements
Impossible. When the function takes the array, all that is really passed is a pointer to its first element. That's how arrays work.
is there a way to pass arrays as a pointer and without passing the size?
Arrays are always passed as a pointer to the first element, unless you pass by reference. If you only have a pointer to the first element, then you need the additional size information somehow, somewhere. If you pass by reference, then you already have the size information, because it's part of the (complete) type.
You should probably just use std::vector, anyway. Raw arrays don't behave "normally" in many ways, std::vector does.
I'm new to C++ and I learned with different tutorials, in one of them I found an example of code:
I have pointed by numbers of lines, that I completely do not understand;
Does this array in array or something like that?
I can understand the second call, but what is the first doing? There is already
"coordinates[blocks[num]]", aren't there? Why need again blocks(i) ?
How do you make this part of the code easier? Did struct with this arrays
don't make easier getting value from arrays?
Thanks in advance!
// Global vars
Struct Rect {
float left;
}
Rectangle *coordinates;
int *blocks;
coordinates = new Rect[25];
blocks = new int[25];
// in method storing values
const int currentBlock = 0; //var in cycle
coordinates[currentBlock].left = column;
blocks[currentBlock] = currentBlock;
//get element method
const Rect& classA::Coords(int num) const
{
return coordinates[blocks[num]]; //(2)
}
//and calling this method like
Coords(blocks[i]); //(3)
Coords(i); //(3)
// (4)
No, not really. Lots of people will think of them as arrays and even describe them as arrays, but they're actually not. coordinates and blocks are both pointers. They just store a single address of a Rect and an int respectively.
However, when you do coordinates = new Rect[25];, for example, you are allocating an array of 25 Rects and setting the pointer coordinates to point at the first element in that array. So, while coordinates itself is a pointer, it's pointing at the first element in an array.
You can index coordinates and blocks like you would an array. For example, coordinates[3] will access the 4th element of the array of Rects you allocated. The reason why this behaves the same as arrays is because it actually is the same. When you have an actual array arr, for example, and you do arr[4], the array first gets converted to a pointer to its first element and then the indexing occurs.
No, this is not an array of arrays. What it is doing is looking up a value in one array (blocks[num]) and using that to index the next array (coordinates[blocks[num]]). So one array is storing indices into the other array.
I'll ignore that this won't compile, but in both cases you are passing an int to the Coords function. The first case looks incorrect, but might not be. It is taking the value at blocks[i], passing that to the function then using that value to index blocks to get another value, then using that other value to index coordinates. In the second case, you are just passing i, which is being used to index blocks to give you a value with which you index coordinates.
That's a broad question that I don't think I can answer without knowing exactly what you want to simplify and without seeing some real valid code.
I've programmed with other languages, but now that I am learning C++, I've found a problem. I am trying to solve a problem with recursion, with a method that takes an array as an argument. I thought about using a public array, maybe, but I can't use the array either way.
From what I've read, it seems to me that it has something to do with the memory. (I thought that, even though it consumes a lot of memory, creating it again with each call would work.)
Here's some code:
static void FindSolution(int row, int column, bool answer[][8][8]) {
for(int i = 0; i < 8; i++)
//Some processing…
bool temp = true;
FindSolution(0, column + 1, answer[row][column] = temp);
}
}
How do I get to actually use the array? Somehow.
The error:
error: array type 'bool [8]' is not assignable
FindSolution(0, column + 1, answer[row][column] = temp);
You have an extra [] on your array. You've declared it as a 3D array, but then you try to assign to it like it is a 2D array. The compiler gets upset because you try to assign a bool value to an array, which is exactly what you are doing:
answer[row][column] = temp;
temp has type bool, but answer[row][column] has type bool[8].
Instead declare the argument without the extra []:
static void FindSolution(int row, int column, bool answer[8][8]) {
You keep incrementing 'column', but you never check it to make sure it doesn't reach 8. When it does reach 8, you're off the end of the array, and you get an error.
There are a few immediate problems with this.
First Problem: Function signature is incorrect
You've declared the third parameter as a 3-dimensional array, but you only want to deal with two dimensions it seems. There are a couple of ways you can redeclare this function to accept a 2D array, for all the options see the accepted answer here. Personally, in this situation I'd go with a template option unless there is a specific reason not to. Something like the following:
template<size_t _rows, size_t _columns>
static void FindSolution(int row, int column, bool (&answer)[_rows][_columns]) {
// todo: Some processing...
}
This allows you to accurately know the size of the array at compile time, of course this won't work so well with dynamically allocated arrays but seeing as you seemed to know the dimensions of the array already at compile time, I figured this wasn't an issue. If it is, check the other ways of passing a 2D array to a function in the link I attached.
Second issue: Recursive call
The second issue is how you're doing your recursive call.
FindSolution(0, column + 1, answer[row][column] = temp);
The result of the assignation of temp to the specific location in the answer array is not the answer array, but rather the value of temp. Effectively the following statement:
answer[row][column] = temp
Is trying to pass a single bool value as a 2-dimensional array, which won't work. In order to correctly call the method again you'll need to do your assignation of temp to the answer array, then call the function again.
answer[row][column] = temp;
FindSolution<_rows,_columns>(0, column + 1, answer);
Should work fine. (Note the explicit template arguments here <_rows,_columns>, this is only needed if you're using the function signature I posted above which made use of templates)