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 };
Related
This question already has answers here:
How do I find the length of an array?
(30 answers)
Closed 8 years ago.
I have this code.
int x[5];
printf("%d\n",sizeof(x) );
int *a;
a = new int[3];
printf("%d\n",sizeof(*a));
When I pass a 'static' array to sizeof(), it returns the dimension of the declared array multiplied by the number of bytes that the datatype uses in memory. However, a dynamic array seems to be different. My question is what should I do to get the size of an 'dynamic' array?
PD: Could it be related to the following?
int *a;
a=new int[3];
a[0]=3;
a[1]=4;
a[2]=5;
a[3]=6;
Why can I modify the third position if it's supposed I put a 'limit' in "a=new int[3]".
When I pass a 'static' array to sizeof(), it returns the dimension of the declared array multiplied by the number of bytes that the datatype uses in memory.
Correct, that is how the size of the entire array is computed.
However, a dynamic array seems to be different.
This is because you are not passing a dynamic array; you are passing a pointer. Pointer is a data type with the size independent of the size of the block of memory to which it may point, hence you always get a constant value. When you allocate memory for your dynamically sized memory block, you need to store the size of allocation for future reference:
size_t count = 123; // <<== You can compute this count dynamically
int *array = new int[count];
cout << "Array size: " << (sizeof(*array) * count) << endl;
C++14 will have variable-length arrays. These arrays will provide a proper size when you check sizeof.
Could it be related to the following? [...]
No, it is unrelated. Your code snippet shows undefined behavior (writing past the end of the allocated block of memory), meaning that your code is invalid. It could crash right away, lead to a crash later on, or exhibit other arbitrary behavior.
In C++ arrays do not have any intrinsic size at runtime.
At compile time one can use sizeof as you showed in order to obtain the size known to the compiler, but if the actual size is not known until runtime then it is the responsibility of the program to keep track of the length.
Two popular strategies are:
Keep a separate variable that contains the current length of the array.
Add an extra element to the end of the array that contains some sort of marker value that indicates that it's the last element. For example, if your array is known to be only of positive integers then you could use -1 as your marker.
If you do not keep track of the end of your array and you write beyond what you allocated then you risk overwriting other data stored adjacent to the array in memory, which could cause crashes or other undefined behavior.
Other languages tend to use the former strategy and provide a mechanism for obtaining the current record of the length. Some languages also allow the array to be dynamically resized after it's created, which usually involves creating a new array and copying over all of the data before destroying the original.
The vector type in the standard library provides an abstraction over arrays that can be more convenient when the size of the data is not known until runtime. It keeps track of the current array size, and allows the array to grow later. For example:
#include <vector>
int main() {
std::vector<int> a;
a.push_back(3);
a.push_back(4);
a.push_back(5);
a.push_back(6);
printf("%d\n", a.size());
return 0;
}
As a side-note, since a.size() (and sizeof(...)) returns a size_t, which isn't necessarily the same size as an int (though it happens to be on some platforms), using printf with %d is not portable. Instead, one can use iostream, which is also more idiomatic C++:
#include <iostream>
std::cout << a.size() << '\n';
You do not, at least not in standard C++. You have to keep track of it yourself, use an alternative to raw pointers such as std::vector that keeps track of the allocated size for you, or use a non-standard function such as _msize https://learn.microsoft.com/en-us/cpp/c-runtime-library/reference/msize?view=msvc-160 on Microsoft Windows or malloc_size https://developer.apple.com/library/archive/documentation/System/Conceptual/ManPages_iPhoneOS/man3/malloc_size.3.html on MacOS X.
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'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.
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]);
void pushSynonyms (string synline, char matrizSinonimos [1024][1024]){
stringstream synstream(synline);
vector<int> synsAux;
int num;
while (synstream >> num) {synsAux.push_back(num);}
int index=0;
while (index<(synsAux.size()-1)){
int primerSinonimo=synsAux[index];
int segundoSinonimo=synsAux[++index];
matrizSinonimos[primerSinonimo][segundoSinonimo]='S';
matrizSinonimos [segundoSinonimo][primerSinonimo]='S';
}
}
and the call..
char matrizSinonimos[1024][1024];
pushSynonyms("1 7", matrizSinonimos)
It's important for me to pass matrizSinonimos by reference.
Edit: took away the & from &matrizSinonimos.
Edit: the runtime error is:
An unhandled win32 exception occurred in program.exe [2488]![alt text][1]
What's wrong with it
The code as you have it there - i can't find a bug. The only problem i spot is that if you provide no number at all, then this part will cause harm:
(synsAux.size()-1)
It will subtract one from 0u . That will wrap around, because size() returns an unsigned integer type. You will end up with a very big value, somewhere around 2^16 or 2^32. You should change the whole while condition to
while ((index+1) < synsAux.size())
You can try looking for a bug around the call side. Often it happens there is a buffer overflow or heap corruption somewhere before that, and the program crashes at a later point in the program as a result of that.
The argument and parameter stuff in it
Concerning the array and how it's passed, i think you do it alright. Although, you still pass the array by value. Maybe you already know it, but i will repeat it. You really pass a pointer to the first element of this array:
char matrizSinonimos[1024][1024];
A 2d array really is an array of arrays. The first lement of that array is an array, and a pointer to it is a pointer to an array. In that case, it is
char (*)[1024]
Even though in the parameter list you said that you accept an array of arrays, the compiler, as always, adjusts that and make it a pointer to the first element of such an array. So in reality, your function has the prototype, after the adjustments of the argument types by the compiler are done:
void pushSynonyms (string synline, char (*matrizSinonimos)[1024]);
Although often suggested, You cannot pass that array as a char**, because the called function needs the size of the inner dimension, to correctly address sub-dimensions at the right offsets. Working with a char** in the called function, and then writing something like matrizSinonimos[0][1], it will try to interpret the first sizeof(char**) characters of that array as a pointer, and will try to dereference a random memory location, then doing that a second time, if it didn't crash in between. Don't do that. It's also not relevant which size you had written in the outer dimension of that array. It rationalized away. Now, it's not really important to pass the array by reference. But if you want to, you have to change the whole thingn to
void pushSynonyms (string synline, char (&matrizSinonimos)[1024][1024]);
Passing by reference does not pass a pointer to the first element: All sizes of all dimensions are preserved, and the array object itself, rather than a value, is passed.
Arrays are passed as pointers - there's no need to do a pass-by-reference to them. If you declare your function to be:
void pushSynonyms(string synline, char matrizSinonimos[][1024]);
Your changes to the array will persist - arrays are never passed by value.
The exception is probably 0xC00000FD, or a stack overflow!
The problem is that you are creating a 1 MB array on the stack, which probably is too big.
try declaring it as:
void pushSynonyms (const string & synline, char *matrizSinonimos[1024] )
I believe that will do what you want to do. The way you have it, as others have said, creates a 1MB array on the stack. Also, changing synline from string to const string & eliminates pushing a full string copy onto the stack.
Also, I'd use some sort of class to encapsulate matrizSinonimos. Something like:
class ms
{
char m_martix[1024][1024];
public:
pushSynonyms( const string & synline );
}
then you don't have to pass it at all.
I'm at a loss for what's wrong with the code above, but if you can't get the array syntax to work, you can always do this:
void pushSynonyms (string synline, char *matrizSinonimos, int rowsize, int colsize )
{
// the code below is equivalent to
// char c = matrizSinonimos[a][b];
char c = matrizSinonimos( a*rowsize + b );
// you could also Assert( a < rowsize && b < colsize );
}
pushSynonyms( "1 7", matrizSinonimos, 1024, 1024 );
You could also replace rowsize and colsize with a #define SYNONYM_ARRAY_DIMENSION 1024 if it's known at compile time, which will make the multiplication step faster.
(edit 1) I forgot to answer your actual question. Well: after you've corrected the code to pass the array in the correct way (no incorrect indirection anymore), it seems most probable to me that you did not check you inputs correctly. You read from a stream, save it into a vector, but you never checked whether all the numbers you get there are actually in the correct range. (end edit 1)
First:
Using raw arrays may not be what you actually want. There are std::vector, or boost::array. The latter one is compile-time fixed-size array like a raw-array, but provides the C++ collection type-defs and methods, which is practical for generic (read: templatized) code.
And, using those classes there may be less confusion about type-safety, pass by reference, by value, or passing a pointer.
Second:
Arrays are passed as pointers, the pointer itself is passed by value.
Third:
You should allocate such big objects on the heap. The overhead of the heap-allocation is in such a case insignificant, and it will reduce the chance of running out of stack-space.
Fourth:
void someFunction(int array[10][10]);
really is:
(edit 2) Thanks to the comments:
void someFunction(int** array);
void someFunction(int (*array)[10]);
Hopefully I didn't screw up elsewhere....
(end edit 2)
The type-information to be a 10x10 array is lost. To get what you've probably meant, you need to write:
void someFunction(int (&array)[10][10]);
This way the compiler can check that on the caller side the array is actually a 10x10 array. You can then call the function like this:
int main() {
int array[10][10] = { 0 };
someFunction(array);
return 0;
}