I am currently working on a program that requires a function to figure out array values and then stores those values in arr1[], which has an unknown size and is calculated within the function. It then passes back the entire array along with the size of the array back to the main function. Using a struct, I returned the size, but I cannot return the array for some reason. Can someone please direct me in the right direction with what I'm doing wrong?
Here is my struct:
struct Arr
{
int size_1;
int arr_1[];
};
And here is part of my function that returns the array, where arr1[] is the array in which I need to return:
Arr smallElement(int arr[], int size)
{
Arr tempArr;
for (int count = 0; count < newSize; count++)
{
tempArr.arr_1[count] = arr1[count];
}
return tempArr;
}
This is what I use in my main function to call the function to print the array, but it just prints 3 random numbers every time:
Arr a;
a = smallElement(array, n);
cout << "The array is: ";
for (int count = 0; count < a.size_1; count++)
{
cout << a.arr_1[count] << " ";
}
Inside a struct, int arr_1[] does not define an actual array. It is actually just a placeholder called a flexible array member. In C, this is used by allocating memory for the fully defined part of the struct plus additional space for the array—you have to manually add space when using malloc or other memory allocation. In standard C++, you should not use this. (It is not part of standard C++. It is an extension adopted by some compilers from C, where it is standard.)
When you declared tempArr, it created just the fully defined portion of the struct. When you assigned values to tempArr.arr_1[count], there was no allocated array there, so the behavior is undefined. When you return tempArr, a copy of the struct is returned, but, again, only a copy of the defined portion.
If the caller knows the size the array will be, you may want to have them pass the array to the function, perhaps by allocating it with new or by using std::vector and passing the vector by reference. If the size is not known by the caller, then you may want to have the function allocate the array and return a pointer to it (in which case the caller is responsible for deleting it later), or you may want to pass an empty (or otherwise available for re-use) std::vector to the function, preferably by reference. Inside the function, it can add more elements to the vector, and the std::vector will manage memory allocation for you.
Related
I have class A, in it I have a dynamic array arr and a dynamic variable s which determines the size of the array. At the start of my main I use the constructor for class A which sets s to 2 and fills the arr with dummy objects. Now when I add objects to the array and sort it, objects seem to disappear.
The size of my array (so variable s) gets doubled when there is not enough space in the array, so no free slot available.
The way I display the array is by using a for loop with the limit i < s. So it should display all objects in the array.
Now when I add more objects and the array gets bigger, the previously "lost" objects start to appear again. So it seems like my array is bigger than the variable s, which dynamically determines the size.
How is that possible?
class A
private:
int *s = new int;
obj *arr = new obj[*s];
//triggers when trying to add an object to the array, but every place from
//arr[0] to arr[s-1] is taken by non dummy object
void A::resize {
*s *= 2;
//for loop that fills code with dummy objects
}
void A::display {
for (int i = 0; i < *s; i++) {
displayobject(i);
}
}
I had a very similar code last year in university and everything worked fine, can't find my mistake.
I would just use a vector, where i can easily use a for loop to go through the real size, but our prof wants us to use a dynamically alocated array.
I have a double pointer Array of a structure:
typedef struct Position{
int x;
int y;
} Position;
Position** array = (Position**)malloc(sizeof(Position*)*10); //10 elements
array[0] = (Position*)malloc(sizeof(Position*));
array[0]->x = 10;
array[0]->y = 5;
Can I calculate the length of set array and if so, how?
The normal way for arrays does not work :
int length = sizeof(<array>)/sizeof(<array>[0]);
Once you have dynamically allocated an array, there is no way of finding out the number of elements in it.
I once heard of some hacky way to obtain the size of a memory block, (msize) which would allegedly allow you to infer the size of the data within the block, but I would advice against any such weird tricks, because they are not covered by the standard, they represent compiler-vendor-specific extensions.
So, the only way to know the size of your array is to keep the size of the array around. Declare a struct, put the array and its length in the struct, and use that instead of the naked array.
As you marked the question as C++, I would suggest that you use std::vector, then, after you "allocated some memory" (or requested some memory to allocated by std::vector constructor or by using push_back, or resize), you can simply get the size back using by using std::vector::size.
typedef struct Position{
int x;
int y;
} Position;
std::vector<Position> array(10);
array[0].x = 10;
array[0].y = 5;
size_t size = array.size(); // will be 10
Having only a pointer to some memory block, you cannot defer the size of this memory block. So you cannot defer the number of elements in it.
For arrays of pointers, however, you could infer the number of elements in it under the following conditions:
make sure that every pointer (except the last one) points to a valid object.
for the last pointer in the array, make sure that it is always NULL.
Then you can derive the length by counting until you reach NULL.
Maybe there are some other similar strategies.
Solely from the pointer itself, however, you cannot derive the number of elements in it.
Old question, but in case someone needs it:
#include <stdio.h>
...
int main()
{
char **double_pointer_char;
...
int length_counter = 0;
while(double_pointer_char[length_counter])
length_counter++;
...
return 0;
}
I'm still quite inexperienced in C++ and i'm trying to write sum code to add numbers precisely. This is a dll plugin for some finite difference software and the code is called several million times during a run. I want to write a function where any number of arguments can be passed in and the sum will be returned. My code looks like:
#include <cstdarg>
double SumFunction(int numArgs, ...){ // this allows me to pass any number
// of arguments to my function.
va_list args;
va_start(args,numArgs); //necessary prerequisites for using cstdarg
double myarray[10];
for (int i = 0; i < numArgs; i++) {
myarray[i] = va_arg(args,double);
} // I imagine this is sloppy code; however i cannot create
// myarray{numArgs] because numArgs is not a const int.
sum(myarray); // The actual method of addition is not relevant here, but
//for more complicated methods, I need to put the summation
// terms in a list.
vector<double> vec(numArgs); // instead, place all values in a vector
for (int i = 0; i < numArgs; i++) {
vec.at(i) = va_arg(args,double);
}
sum(vec); //This would be passed by reference, of course. The function sum
// doesn't actually exist, it would all be contained within the
// current function. This is method is twice as slow as placing
//all the values in the static array.
double *vec;
vec = new double[numArgs];
for (int i = 0; i < (numArgs); i++) {
vec[i] = va_arg(args,double);
}
sum(vec); // Again half of the speed of using a standard array and
// increasing in magnitude for every extra dynamic array!
delete[] vec;
va_end(args);
}
So the problem I have is that using an oversized static array is sloppy programming, but using either a vector or a dynamic array slows the program down considerably. So I really don't know what to do. Can anyone help, please?
One way to speed the code up (at the cost of making it more complicated) is to reuse a dynamic array or vector between calls, then you will avoid incurring the overhead of memory allocation and deallocation each time you call the function.
For example declare these variables outside your function either as global variables or as member variables inside some class. I'll just make them globals for ease of explanation:
double* sumArray = NULL;
int sumArraySize = 0;
In your SumFunction, check if the array exists and if not allocate it, and resize if necessary:
double SumFunction(int numArgs, ...){ // this allows me to pass any number
// of arguments to my function.
va_list args;
va_start(args,numArgs); //necessary prerequisites for using cstdarg
// if the array has already been allocated, check if it is large enough and delete if not:
if((sumArray != NULL) && (numArgs > sumArraySize))
{
delete[] sumArray;
sumArray = NULL;
}
// allocate the array, but only if necessary:
if(sumArray == NULL)
{
sumArray = new double[numArgs];
sumArraySize = numArgs;
}
double *vec = sumArray; // set to your array, reusable between calls
for (int i = 0; i < (numArgs); i++) {
vec[i] = va_arg(args,double);
}
sum(vec, numArgs); // you will need to pass the array size
va_end(args);
// note no array deallocation
}
The catch is that you need to remember to deallocate the array at some point by calling a function similar to this (like I said, you pay for speed with extra complexity):
void freeSumArray()
{
if(sumArray != NULL)
{
delete[] sumArray;
sumArray = NULL;
sumArraySize = 0;
}
}
You can take a similar (and simpler/cleaner) approach with a vector, allocate it the first time if it doesn't already exist, or call resize() on it with numArgs if it does.
When using a std::vector the optimizer must consider that relocation is possible and this introduces an extra indirection.
In other words the code for
v[index] += value;
where v is for example a std::vector<int> is expanded to
int *p = v._begin + index;
*p += value;
i.e. from vector you need first to get the field _begin (that contains where the content starts in memory), then apply the index, and then dereference to get the value and mutate it.
If the code performing the computation on the elements of the vector in a loop calls any unknown non-inlined code, the optimizer is forced to assume that unknown code may mutate the _begin field of the vector and this will require doing the two-steps indirection for each element.
(NOTE: that the vector is passed with a cost std::vector<T>& reference is totally irrelevant: a const reference doesn't mean that the vector is const but simply puts a limitation on what operations are permitted using that reference; external code could have a non-const reference to access the vector and constness can also be legally casted away... constness of references is basically ignored by the optimizer).
One way to remove this extra lookup (if you know that the vector is not being resized during the computation) is to cache this address in a local and use that instead of the vector operator [] to access the element:
int *p = &v[0];
for (int i=0,n=v.size(); i<n; i++) {
/// use p[i] instead of v[i]
}
This will generate code that is almost as efficient as a static array because, given that the address of p is not published, nothing in the body of the loop can change it and the value p can be assumed constant (something that cannot be done for v._begin as the optimizer cannot know if someone else knows the address of _begin).
I'm saying "almost" because a static array only requires indexing, while using a dynamically allocated area requires "base + indexing" access; most CPUs however provide this kind of memory access at no extra cost. Moreover if you're processing elements in sequence the indexing addressing becomes just a sequential memory access but only if you can assume the start address constant (i.e. not in the case of std::vector<T>::operator[]).
Assuming that the "max storage ever needed" is in the order of 10-50, I'd say using a local array is perfectly fine.
Using vector<T> will use 3 * sizeof(*T) (at least) to track the contents of the vector. So if we compare that to an array of double arr[10];, then that's 7 elements more on the stack of equal size (or 8.5 in 32-bit build). But you also need a call to new, which takes a size argument. So that takes up AT LEAST one, more likely 2-3 elements of stackspace, and the implementation of new is quite possibly not straightforward, so further calls are needed, which take up further stack-space.
If you "don't know" the number of elements, and need to cope with quite large numbers of elements, then using a hybrid solution, where you have a small stack-based local array, and if numargs > small_size use vector, and then pass vec.data() to the function sum.
I am basically looking for some sort of "dynamic" way of passing the size/length of an array to a function.
I have tried:
void printArray(int arrayName[])
{
for(int i = 0 ; i < sizeof(arrayName); ++i)
{
cout << arrayName[i] << ' ';
}
}
But I realized it only considers its bytesize and not how many elements are on the array.
And also:
void printArray(int *arrayName)
{
while (*arrayName)
{
cout << *arrayName << ' ';
*arrayName++;
}
}
This has at least printed me everything but more than what I expected, so it doesn't actually work how I want it to.
I reckon it is because I don't exactly tell it how big I need it to be so it plays it "safe" and throws me some big size and eventually starts printing me very odd integers after my last element in the array.
So I finally got this work around, yet I believe there is something better out there!:
void printArray(int *arrayName)
{
while (*arrayName)
{
if (*arrayName == -858993460)
{
break;
}
cout << *arrayName << ' ';
*arrayName++;
}
cout << '\n';
}
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460, so I made it break the while loop once this value is encountered.
include <iostream>
include <conio.h>
using namespace std;
// functions prototypes
void printArray (int arrayName[], int lengthArray);
// global variables
//main
int main ()
{
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray,3);
printArray (secondArray,5);
// end of program
_getch();
return 0;
}
// functions definitions
void printArray(int arrayName[], int lengthArray)
{
for (int i=0; i<lengthArray; i++)
{
cout << arrayName[i] << " ";
}
cout << "\n";
}
Thank you very much.
TL;DR answer: use std::vector.
But I realized it [sizeof()] only considers its bytesize and not how many elements are on the array.
That wouldn't be a problem in itself: you could still get the size of the array using sizeof(array) / sizeof(array[0]), but the problem is that when passed to a function, arrays decay into a pointer to their first element, so all you can get is sizeof(T *) (T being the type of an element in the array).
About *arrayName++:
This has at least printed me everything but more than what I expected
I don't even understand what inspired you to calculate the size of the array in this way. All that this code does is incrementing the first object in the array until it's zero.
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460
That's a terrible assumption and it also relies on undefined behavior. You can't really be sure what's in the memory after the first element of your array, you should not even be accessing it.
Basically, in C++, if you want to know the size of a raw array from within a function, then you have to keep track of it manually (e. g. adding an extra size_t size argument), because of the way arrays are passed to functions (remember, they "decay into" a pointer). If you want something more flexible, consider using std::vector<int> (or whatever type of objects you want to store) from the C++ standard library -- it has a size() method, which does exactly what you want.
1st try
When arrays are passed into functions they decay to pointers. Normally, using sizeof on an array would give you its size in bytes which you could then divide by the size in bytes of each element and get the number of elements. But now, since you have a pointer instead of an array, calling sizeof just gives you the size of the pointer (usually 4 or 8 bytes), not the array itself and that's why this fails.
2nd try
The while loop in this example assumes that your array ends with a zero and that's very bad (unless you really did use a zero as a terminator like null-terminated strings for example do). If your array doesn't end with a zero you might be accessing memory that isn't yours and therefore invoking undefined behavior. Another thing that could happen is that your array has a zero element in the middle which would then only print the first few elements.
3rd try
This special value you found lurking at the end of your array can change any time. This value just happened to be there at this point and it might be different another time so hardcoding it like this is very dangerous because again, you could end up accessing memory that isn't yours.
Your final code
This code is correct and passing the length of the array along with the array itself is something commonly done (especially in APIs written in C). This code shouldn't cause any problems as long as you don't pass a length that's actually bigger than the real length of the array and this can happen sometimes so it is also error prone.
Another solution
Another solution would be to use std::vector, a container which along with keeping track of its size, also allows you to add as many elements as you want, i.e. the size doesn't need to be known at runtime. So you could do something like this:
#include <iostream>
#include <vector>
#include <cstddef>
void print_vec(const std::vector<int>& v)
{
std::size_t len = v.size();
for (std::size_t i = 0; i < len; ++i)
{
std::cout << v[i] << std::endl;
}
}
int main()
{
std::vector<int> elements;
elements.push_back(5);
elements.push_back(4);
elements.push_back(3);
elements.push_back(2);
elements.push_back(1);
print_vec(elements);
return 0;
}
Useful links worth checking out
Undefined behavior: Undefined, unspecified and implementation-defined behavior
Array decay: What is array decaying?
std::vector: http://en.cppreference.com/w/cpp/container/vector
As all the other answers say, you should use std::vector or, as you already did, pass the number of elements of the array to the printing function.
Another way to do is is by putting a sentinel element (a value you are sure it won't be inside the array) at the end of the array. In the printing function you then cycle through the elements and when you find the sentinel you stop.
A possible solution: you can use a template to deduce the array length:
template <typename T, int N>
int array_length(T (&array)[N]) {
return N;
}
Note that you have to do this before the array decays to a pointer, but you can use the technique directly or in a wrapper.
For example, if you don't mind rolling your own array wrapper:
template <typename T>
struct array {
T *a_;
int n_;
template <int N> array(T (&a)[N]) : a_(a), n_(N) {}
};
You can do this:
void printArray(array<int> a)
{
for (int i = 0 ; i < a.n_; ++i)
cout << a.a_[i] << ' ';
}
and call it like
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray);
printArray (secondArray);
The key is that the templated constructor isn't explicit so your array can be converted to an instance, capturing the size, before decaying to a pointer.
NB. The wrapper shown isn't suitable for owning dynamically-sized arrays, only for handling statically-sized arrays conveniently. It's also missing various operators and a default constructor, for brevity. In general, prefer std::vector or std::array instead for general use.
... OP's own attempts are completely addressed elsewhere ...
Using the -858993460 value is highly unreliable and, in fact, incorrect.
You can pass a length of array in two ways: pass an additional parameter (say size_t length) to your function, or put a special value to the end of array. The first way is preferred, but the second is used, for example, for passing strings by char*.
In C/C++ it's not possible to know the size of an array at runtime. You might consider using an std::vector class if you need that, and it has other advantages as well.
When you pass the length of the array to printArray, you can use sizeof(array) / sizeof(array[0]), which is to say the size in bytes of the whole array divided by the size in bytes of a single element gives you the size in elements of the array itself.
More to the point, in C++ you may find it to your advantage to learn about std::vector and std::array and prefer these over raw arrays—unless of course you’re doing a homework assignment that requires you to learn about raw arrays. The size() member function will give you the number of elements in a vector.
In C/C++, native arrays degrade to pointers as soon as they are passed to functions. As such, the "length" parameter has to be passed as a parameter for the function.
C++ offers the std::vector collection class. Make sure when you pass it to a function, you pass it by reference or by pointer (to avoid making a copy of the array as it's passed).
#include <vector>
#include <string>
void printArray(std::vector<std::string> &arrayName)
{
size_t length = arrayName.size();
for(size_t i = 0 ; i < length; ++i)
{
cout << arrayName[i] << ' ';
}
}
int main()
{
std::vector<std::string> arrayOfNames;
arrayOfNames.push_back(std::string("Stack"));
arrayOfNames.push_back(std::string("Overflow"));
printArray(arrayOfNames);
...
}
I have an array that is being dynamically created based on user input, and I need to pass that array to a function where it will be manipulated and the results passed back out to the original array. When I try to use
void MyFunction(int (&MyArray)[])
my compiler complains that it doesn't know how big the array is.
You can't. You could use a std::vector though.
You can give a pointer to the first element of the array + a second parameter holding the size of the array.
If it's just an array, why not pass the array itself and its size as a second parameter? (by passing the array as an int* or int[], same thing as far as C++ is concerned).
As the value of the variable containing your array is only the pointer to the first element of your array, you don't end up killing your runtime by copying the contents of the array, but just a pointer which is as small as you can get in this case.
void MyFunction( int MyArray[], int size ) { /* edit the array */ }
int main() {
// read nrElements
// ...
// create the array
int *a = new int[nrElements];
// populate it with data
// ...
// and then
MyFunction(a, nrElements);
}
You should use a std::vector only if you want to resize the array in your function (e.g. add new elements to it), but otherwise you can just stick to this approach because it's faster.
Btw, the only case you would need a reference to an array is if you want to change the value of the variable you pass in when you call the function, as in, make it point to a different array. Like in:
void createMyArray(int* &array, int nrElements) {
array = new int[nrElements];
for (int i = 0; i < nrElements; ++i) {
array[i] = 0;
}
}
int *a = (int []) {1, 2, 3};
std::cout << a[0] << std::endl; // prints 1
createMyArray(a, 10);
// now a points to the first element of a 10-element array
std::cout << a[0] << std::endl; // prints 0
But you said the array is already created before providing it to the function, there's no point in using references.