I'm trying to do a little application that would calculate some paths for a given graph.
I've created a class to handle simple graphs, as follows:
class SimpleGraph {
int _nbNodes;
int _nbLines;
protected:
int AdjMatrix[_nbNodes, _nbNodes]; //Error happens here...
int IncMatrix[_nbNodes, _nbLines]; //...and here!
public:
SimpleGraph(int nbNodes, int nbLines) { this->_nbNodes = nbNodes - 1; this->_nbLines = nbLines - 1; };
virtual bool isSimple();
};
At compilation time, I get an error on the two protected members declaration.
I don't understand what is wrong, as there is only one constructor that takes these values as parameters. As such, they cannot be uninitialized.
What am I missing here?
The compiler needs to know how much space to allocate for a member of class SimpleGraph. However, since AdjMatrix and IncMatrix are defined on the stack and their sizes are determined at run-time (i.e., after compilation), it cannot do that. Specifically, the standard says that the size of an array in a class must be a constexpr.
To fix this, you can:
Allocate AdjMatrix and IncMatrix on the heap instead and then you can allocate memory at runtime.
Use a fixed size for the two arrays and keep them on the stack.
--
Another major issue with your code is that you cannot create multi-dimensional arrays using a comma (AdjMatrix[int, int]). You must instead either use:
AdjMatrix[int][int]
AdjMatrix[int * int]
Objects in C++ have a fixed size that needs to be known at compilation time. The size of AdjMatrix and InMatrix are not known at compilation time, only at run time.
In the lines
int AdjMatrix[_nbNodes, _nbNodes]; //Error happens here...
int IncMatrix[_nbNodes, _nbLines]; //...and here!
The array notation is wrong. You cannot specify a 2 dimensional array that way in C++. The correct notation uses brackets on each dimension, as for instance:
int data[5][2];
Regarding the problem you are facing, the dimensions of an array in C++ must be specified at compile time, ie. the compiler must know what are the values used to indicate the array dimension when compiling the program. This is clearly not the case here. You must revert to use integer literals, as in my example, or change the code to use vectors:
std::vector<std::vector<int> > AdjMatrix;
and in the constructor:
SimpleGraph(int nbNodes, int nbLines) : AdjMatrix(nbNodes) {
for (int i = 0; i< nbNodes; i++)
AdjMatrix[i].resize(20);
}
Note that you won't need _nbNodes anymore, and use instead the size() method on AdjMatrix. You will have to do the same for IncMatrix.
Another option, if you know the values at compile time, is to use macros to define them symbolically.
#define NBNODES 20
int AdjMatrix[NBNODES][NBNODES];
but since you wish to pass them as constructor parameter, this may not fit your need. Still, if you know that the parameters are constants at compile time, you might be able use the C++11 constexpr qualifier on the constructor parameters.
Related
I understand that this question was asked before but I don't get why it doesn't work in my case
void calc(vector<char> zodis1, vector<char> zodis2, vector<char> zodisAts,int zo1,int zo2,int zoA)
{
int i,u=0;
int zod1[zo1]=0;
int zod2[zo2]=0;
int zodA[zoA]=0;
}
All 3 of zod1, zod2, zoA gives me error: variable-sized object may not be initialized c++
But compiler should know the meaning of zo before initialization cause cout<<zo1; works and print out the meaning
So whats the problem?
You can declare an array only with constant size, which can be deduced at compile time. zo1,zo2 and zoA are variables, and the values can be known only at runtime.
To elaborate, when you allocate memory on the stack, the size must be known at compile time. Since the arrays are local to the method, they will be placed on the stack. You can either use constant value, or allocate memory in the heap using new, and deallocate when done using delete, like
int* zod1 = new int[zo1];
//.... other code
delete[] zod1;
But you can use vector instead of array here also, and vector will take care of allocation on the heap.
As a side note, you should not pass vector by value, as the whole vector will be copied and passed as argument, and no change will be visible at the caller side. Use vector<char>& zodis1 instead.
Here is the fix, you can write the following lines instead of the line where you got the error;
Alternative 1 you can use vectors:
vector<int> zod1(zo1, 0);
Alternative 2 (for example, since w know "0 <= s.length <= 100", we can use constant value):
int zod1[100] = { 0 };
What is the difference between this two array definitions and which one is more correct and why?
#include <stdio.h>
#define SIZE 20
int main() {
// definition method 1:
int a[SIZE];
// end definition method 1.
// defintion method 2:
int n;
scanf("%d", &n);
int b[n];
// end definition method 2.
return 0;
}
I know if we read size, variable n, from stdin, it's more correct to define our (block of memory we'll be using) array as a pointer and use stdlib.h and array = malloc(n * sizeof(int)), rather than decalring it as int array[n], but again why?
It's not "more correct" or "less correct". It either is xor isn't correct. In particular, this works in C, but not in C++.
You are declaring dynamic arrays. Better way to declare Dynamic arrays as
int *arr; // int * type is just for simplicity
arr = malloc(n*sizeof(int*));
this is because variable length arrays are only allowed in C99 and you can't use this in c89/90.
In (pre-C99) C and C++, all types are statically sized. This means that arrays must be declared with a size that is both constant and known to the compiler.
Now, many C++ compilers offer dynamically sized arrays as a nonstandard extension, and C99 explicitly permits them. So int b[n] will most likely work if you try it. But in some cases, it will not, and the compiler is not wrong in those cases.
If you know SIZE at compile-time:
int ar[SIZE];
If you don't:
std::vector<int> ar;
I don't want to see malloc anywhere in your C++ code. However, you are fundamentally correct and for C that's just what you'd do:
int* ptr = malloc(sizeof(int) * SIZE);
/* ... */
free(ptr);
Variable-length arrays are a GCC extension that allow you to do:
int ar[n];
but I've had issues where VLAs were disabled but GCC didn't successfully detect that I was trying to use them. Chaos ensues. Just avoid it.
Q1 : First definition is the static array declaration. Perfectly correct.
It is when you have the size known, so no comparison with VLA or malloc().
Q2 : Which is better when taking size as an input from the user : VLA or malloc .
VLA : They are limited by the environment's bounds on the size of automatic
allocation. And automatic variables are usually allocated on the stack which is relatively
small.The limitation is platform specific.Also, this is in c99 and above only.Some ease of use while declaring multidimensional arrays is obtained by VLA.
Malloc : Allocates from the heap.So, for large size is definitely better.For, multidimensional arrays pointers are involved so a bit complex implementataion.
Check http://bytes.com/topic/c/answers/578354-vla-feature-c99-vs-malloc
I think that metod1 could be little bit faster, but both of them are correct in C.
In C++ first will work, but if you want to use a second you should use:
int size = 5;
int * array = new int[size];
and remember to delete it:
delete [] array;
I think it gives you more option to use while coding.
If you use malloc or other dynamic allocation to get a pointer. You will use like p+n..., but if you use array, you could use array[n]. Also, while define pointer, you need to free it; but array does not need to free.
And in C++, we could define user-defined class to do such things, and in STL, there is std::vector which do the array-things, and much more.
Both are correct. the declaration you use depends on your code.
The first declaration i.e. int a[size]; creates an array with a fixed size of 20 elements.
It is helpful when you know the exact size of the array that will be used in the code. for example, you are generating
table of a number n up till its 20th multiple.
The second declaration allows you to make an array of the size that you desire.
It is helpful when you will need an array of different sizes, each time the code is executed for example, you want to generate the fibonacci series till n. In that case, the size of the array must be n for each value of n. So say you have n = 5, in this case int a [20] will waste memory because only the first five slots will be used for the fibonacci series and the rest will be empty. Similarly if n = 25 then your array int a[20] will become too small.
The difference if you define array using malloc is that, you can pass the size of array dynamically i.e at run time. You input a value your program has during run time.
One more difference is that arrays created using malloc are allocated space on heap. So they are preserved across function calls unlike static arrays.
example-
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
int *a;
scanf("%d",&n);
a=(int *)malloc(n*sizeof(int));
return 0;
}
I am getting the error
"The expression needs to be a constant"
when I try to do this:
float mat1[m_Floats.size()][iNumClass];
Can I trick the compiler (VS2010) into compiling this anyway?
No. Only C99 specify dynamic array allocation (i.e. where size is known only at compile time). Maybe there is a MSVC extension to the standard, but you should go the canonical way in creating the array of pointers and each float subarray using new, like in:
float **mat1 = new float*[m_Floats.size()];
for (int i = 0; i < m_Floats.size(); ++i) {
mat1[i] = new float[iNumClass];
}
well, instead of "tricking the compiler", you can dynamically allocate your matrix with the operator new
No. The size of a declared array needs to be known at compile time. The value of m_Floats.size() depends on how many members are in that object.
If you need to allocate arrays with variable size, you'll need to handle it yourself with new or some appropriate classes/methods that encapsulate that for you.
I want to pass a reference to an array from one object GameModel to another PersonModel, store reference and then work with this array inside PersonModel just like inside GameModel, but...
I have a terrible misunderstanding of passing an array process: In the class PersonModel I want to pass an array by reference in a constructor (see code block below). But the marked line throws the compile error
PersonModel::PersonModel( int path[FieldSize::HEIGHT][FieldSize::WIDTH], int permissionLevel ) {
this->path = path; //<------ ERROR here:
//PersonModel.cpp:14:22: error: incompatible types in assignment of 'int (*)[30]' to 'int [31][30]'
this->permissionLevel = permissionLevel;
}
Here is the header file PersonModel.h
#ifndef PERSON_MODEL
#define PERSON_MODEL
#include "data/FieldSize.h"
namespace game{
class IntPosition;
class MotionDirection;
class PersonModel {
protected:
int path[FieldSize::HEIGHT][FieldSize::WIDTH];
int permissionLevel;
public:
PersonModel( int path[FieldSize::HEIGHT][FieldSize::WIDTH], int permissionLevel );
void setMotionDirection ( MotionDirection* md);
void step(long time);
void reset(long time);
};
}
#endif
As I see now, I can change the int path[FieldSize::HEIGHT][FieldSize::WIDTH]; declaration to int (*path)[FieldSize::WIDTH]; but it is much more confusing.
Help me understand this topic: what is the proper way to store the passed reference to an array to work with it later, like with usual 2D array.
UPDATE:
This array is a map of game field tiles properties represented by bit-masks, so it is read-only actually. All the incapsulated objects of GameModel class should read this array, but I definitely don't want to duplicate it or add some extra functionality.
There are no frameworks just bare Android-NDK.
I think you've fallen into the classic trap of believing someone who's told you that "arrays and pointers are the same in C".
The first thing I'd do would be to define a type for the array:
typedef int PathArray[FieldSize::HEIGHT][FieldSize::WIDTH];
You then don't need to worry about confusions between reference to array of ints vs array of references to ints.
Your PersonModel then contains a reference to one of these.
PathArray &path;
and, because its a reference it must be initialised in the constructors initialization list rather than in the constructor body.
PersonModel::PersonModel( PathArray &aPath, int aPermissionLevel ) :
path(aPath),
permissionLevel(aPermissionLevel)
{
}
Of course, holding references like this is a little scary so you might want to consider using a boost::shared_ptr or something similar instead to make the lifetime management more robust.
You cannot assign arrays as you do with value types in C++
int path[x][y] resolves to the type int (*)[y]
Possible solutions are:
Using memcpy/copy
Using std::array
You can't assign to an array like that. However you can use the fact that an array is a contiguous memory area, even when having an array of arrays, and use e.g. memcpy to copy the array:
memcpy(this->path, path, FieldSize::HEIGHT * FieldSize::WIDTH * sizeof(int));
You would have to pass a pointer to the 2d-array as you cannot pass the array as you have stated in the code snippet.
I would suggest using the STL array type. Admittedly std::array is C++ '11 standard and therefore old compiler may not support it. You can also use vector which has been around longer.
vector<vector<int>>path;
You will have to resize the 2d-vector in the constructor.
Indexing would look a bit funny:
path[1].[1] ....
With vectors, you can then pass it by reference.
the name of the array is a pointer on first element
so,
you can try
PersonModel( int (*path)[FieldSize::HEIGHT][FieldSize::WIDTH], int permissionLevel );
In C++ '=' implemented for primitive types like int and double but not for array(array is not a primitive type), so you should never use '=' to assign an array to new array, instead you should use something as memcpy to copy array. memcpy copy a memory over another memory, so you can use it to copy an array over another array:
// memcpy( dst, src, size );
memcpy( this->path, path, FieldSize::HEIGHT * FieldSize * WEIGHT * sizeof(int) );
Okay so you have and array A[]... that is passed to you in some function say with the following function prototype:
void foo(int A[]);
Okay, as you know it's kind of hard to find the size of that array without knowing some sort of ending variable or knowing the size already...
Well here is the deal though. I have seem some people figure it out on a challenge problem, and I don't understand how they did it. I wasn't able to see their source code of course, that is why I am here asking.
Does anyone know how it would even be remotely possible to find the size of that array?? Maybe something like what the free() function does in C??
What do you think of this??
template<typename E, int size>
int ArrLength(E(&)[size]){return size;}
void main()
{
int arr[17];
int sizeofArray = ArrLength(arr);
}
The signature of that function is not that of a function taking an array, but rather a pointer to int. You cannot obtain the size of the array within the function, and will have to pass it as an extra argument to the function.
If you are allowed to change the signature of the function there are different alternatives:
C/C++ (simple):
void f( int *data, int size ); // function
f( array, sizeof array/sizeof array[0] ); // caller code
C++:
template <int N>
void f( int (&array)[N] ); // Inside f, size N embedded in type
f( array ); // caller code
C++ (though a dispatch):
template <int N>
void f( int (&array)[N] ) { // Dispatcher
f( array, N );
}
void f( int *array, int size ); // Actual function, as per option 1
f( array ); // Compiler processes the type as per 2
You cannot do that. Either you have a convention to signal the end of the array (e.g. that it is made of non-zero integers followed by a 0), or you transmit the size of the array (usually as an additional argument).
If you use the Boehm garbage collector (which has a lot of benefit, in particular you allocate with GC_malloc and friends but you don't care about free-ing memory explicitly), you could use the GC_size function to give you the size of a GC_malloc-ed memory zone, but standard malloc don't have this feature.
You're asking what we think of the following code:
template<typename E, int size>
int ArrLength(E(&)[size]){return size;}
void main()
{
int arr[17];
int sizeofArray = ArrLength(arr);
}
Well, void main has never been standard, neither in C nor in C++.
It's int main.
Regarding the ArrLength function, a proper implementation does not work for local types in C++98. It does work for local types by C++11 rules. But in C++11 you can write just end(a) - begin(a).
The implementation you show is not proper: it should absolutely not have int template argument. Make that a ptrdiff_t. For example, in 64-bit Windows the type int is still 32-bit.
Finally, as general advice:
Use std::vector and std::array.
One relevant benefit of this approach is that it avoid throwing away the size information, i.e. it avoids creating the problem you're asking about. There are also many other advantages. So, try it.
The first element could be a count, or the last element could be a sentinel. That's about all I can think of that could work portably.
In new code, for container-agnostic code prefer passing two iterators (or pointers in C) as a much better solution than just passing a raw array. For container-specific code use the C++ containers like vector.
No you can't. Your prototype is equivalent to
void foo(int * A);
there is obviously no size information. Also implementation dependent tricks can't help:
the array variable can be allocated on the stack or be static, so there is no information provided by malloc or friends
if allocated on the heap, a user of that function is not forced to call it with the first element of an allocation.
e.g the following are valid
int B[22];
foo(B);
int * A = new int[33];
foo(A + 25);
This is something that I would not suggest doing, however if you know the address of the beginning of the array and the address of the next variable/structure defined, you could subtract the address. Probably not a good idea though.
Probably an array allocated at compile time has information on its size in the debug information of the executable. Moreover one could search in the code for all the address corresponding to compile time allocated variables and assume the size of the array is minus the difference between its starting address and the next closest starting address of any variable.
For a dinamically allocated variable it should be possible to get its size from the heap data structures.
It is hacky and system dependant, but it is still a possible solution.
One estimate is as follows: if you have for instance an array of ints but know that they are between (stupid example) 0..80000, the first array element that's either negative or larger than 80000 is potentially right past the end of the array.
This can sometimes work because the memory right past the end of the array (I'm assuming it was dynamically allocated) won't have been initialized by the program (and thus might contain garbage values), but might still be part of the allocated pages, depending on the size of the array. In other cases it will crash or fail to provide meaningful output.
All of the other answers are probably better, i.e. you either have to pass the length of the array or terminate it with a special byte sequence.
The following method is not portable, but it works for me in VS2005:
int getSizeOfArray( int* ptr )
{
int size = 0;
void* ptrToStruct = ptr;
long adr = (long)ptrToStruct;
adr = adr - 0x10;
void* ptrToSize = (void*)adr;
size = *(int*)ptrToSize;
size /= sizeof(int);
return size;
}
This is entirely dependent of the memory model of your compiler and system so, again, it is not portable. I bet there are equivalent methods for other platforms. I would never use this in a production environment, merely stating this as an alternative.
You can use this: int n = sizeof(A) / sizeof(A[0]);