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.
Related
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
I'm learning C++ from a course on Udacity.
Can you explain to me why setGrades() was defined as a pass-by-pointer-to-value function? Why is there an error with passing by value? In the code below, I omitted the definition for printGrades() and setID().
#include<iostream>
using namespace std;
const int SIZE = 5;
template <class T>
class StudentRecord
{
private:
const int size = SIZE;
T grades[SIZE];
int studentId;
public:
StudentRecord(T defaultInput);
void setGrades(T* input);
void setId(int idIn);
void printGrades();
};
template<class T>
StudentRecord<T>::StudentRecord(T defaultInput)
{
for(int i=0; i<SIZE; ++i)
grades[i] = defaultInput;
}
template<class T>
void StudentRecord<T>::setGrades(T* input)
{
for(int i=0; i<SIZE;++i)
{
grades[i] = input[i];
}
}
int main()
{
StudentRecord<int> srInt(-1);
srInt.setId(111111);
int arrayInt[SIZE]={4,3,2,1,4};
srInt.setGrades(arrayInt);
srInt.printGrades();
return 0;
}
The output is supposed to be:
ID# 111111: 4
3
2
1
4
C++ does not allow passing builtin C-style arrays by value. One can pass an array by reference or pass a pointer to the first element of the array. Given such pointer, the entire array can be accessed.
Passing-by-pointer is not a usual term in the literature but people keep coining similar terms time and again, which shows some kind of genuine need. The idea behind the term is as follows: one passes a pointer by value but the goal is to let the function access the pointed-to object as an lvalue (which is normally achieved by passing that object by reference).
That's simply because it is an array you want to give to setGrades because you want to set all SIZE values of the array 'grades'.
As you know perhaps, the name of an array can be used like a pointer to the first value of that array.
For example you could write *arrayInt as a term which is equivalent to arrayInt[0].
So when you pass an pointer to the first element of the array to setGrades, the function can get the other elements of that array with arrayName[i] where i is in between 0 and SIZE.
You want to pass an array to setGrades, you can also define it like this:
void StudentRecord<T>::setGrades(T input[])
However the compiler will convert it to a T* pointer automatically.
Functions will always make its own copy of all parameters you gave during compilation (see: call by value). Right here it's an array, but the compiler can't assign an array to another immediately. The only way to do array assignment is to assign its element one by one (or copy the entire memory chunk of the array), and compiler won't do that. The compiler do know that you can access the entire array if you got a pointer of type T pointed to the first element of the array. It's the alternative way of passing the entire array.
And that is the most common way to pass an array to a function.
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.
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.