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I am fairly new to C++ and have been avoiding pointers. From what I've read online I cannot return an array but I can return a pointer to it. I made a small code to test it and was wondering if this was the normal / correct way to do this:
#include <iostream>
using namespace std;
int* test (int in[5]) {
int* out = in;
return out;
}
int main() {
int arr[5] = {1, 2, 3, 4, 5};
int* pArr = test(arr);
for (int i = 0; i < 5; i++) cout<<pArr[i]<<endl;
cout<<endl;
return 0;
}
Edit: This seems to be no good. How should I rewrite it?
int* test (int a[5], int b[5]) {
int c[5];
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
int* out = c;
return out;
}
Your code as it stands is correct but I am having a hard time figuring out how it could/would be used in a real world scenario. With that said, please be aware of a few caveats when returning pointers from functions:
When you create an array with syntax int arr[5];, it's allocated on the stack and is local to the function.
C++ allows you to return a pointer to this array, but it is undefined behavior to use the memory pointed to by this pointer outside of its local scope. Read this great answer using a real world analogy to get a much clear understanding than what I could ever explain.
You can still use the array outside the scope if you can guarantee that memory of the array has not be purged. In your case this is true when you pass arr to test().
If you want to pass around pointers to a dynamically allocated array without worrying about memory leaks, you should do some reading on std::unique_ptr/std::shared_ptr<>.
Edit - to answer the use-case of matrix multiplication
You have two options. The naive way is to use std::unique_ptr/std::shared_ptr<>. The Modern C++ way is to have a Matrix class where you overload operator * and you absolutely must use the new rvalue references if you want to avoid copying the result of the multiplication to get it out of the function. In addition to having your copy constructor, operator = and destructor, you also need to have move constructor and move assignment operator. Go through the questions and answers of this search to gain more insight on how to achieve this.
Edit 2 - answer to appended question
int* test (int a[5], int b[5]) {
int *c = new int[5];
for (int i = 0; i < 5; i++)
c[i] = a[i]+b[i];
return c;
}
If you are using this as int *res = test(a,b);, then sometime later in your code, you should call delete []res to free the memory allocated in the test() function. You see now the problem is it is extremely hard to manually keep track of when to make the call to delete. Hence the approaches on how to deal with it where outlined in the answer.
Your code is OK. Note though that if you return a pointer to an array, and that array goes out of scope, you should not use that pointer anymore. Example:
int* test (void)
{
int out[5];
return out;
}
The above will never work, because out does not exist anymore when test() returns. The returned pointer must not be used anymore. If you do use it, you will be reading/writing to memory you shouldn't.
In your original code, the arr array goes out of scope when main() returns. Obviously that's no problem, since returning from main() also means that your program is terminating.
If you want something that will stick around and cannot go out of scope, you should allocate it with new:
int* test (void)
{
int* out = new int[5];
return out;
}
The returned pointer will always be valid. Remember do delete it again when you're done with it though, using delete[]:
int* array = test();
// ...
// Done with the array.
delete[] array;
Deleting it is the only way to reclaim the memory it uses.
New answer to new question:
You cannot return pointer to automatic variable (int c[5]) from the function. Automatic variable ends its lifetime with return enclosing block (function in this case) - so you are returning pointer to not existing array.
Either make your variable dynamic:
int* test (int a[5], int b[5]) {
int* c = new int[5];
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
return c;
}
Or change your implementation to use std::array:
std::array<int,5> test (const std::array<int,5>& a, const std::array<int,5>& b)
{
std::array<int,5> c;
for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
return c;
}
In case your compiler does not provide std::array you can replace it with simple struct containing an array:
struct array_int_5 {
int data[5];
int& operator [](int i) { return data[i]; }
int operator const [](int i) { return data[i]; }
};
Old answer to old question:
Your code is correct, and ... hmm, well, ... useless. Since arrays can be assigned to pointers without extra function (note that you are already using this in your function):
int arr[5] = {1, 2, 3, 4, 5};
//int* pArr = test(arr);
int* pArr = arr;
Morever signature of your function:
int* test (int in[5])
Is equivalent to:
int* test (int* in)
So you see it makes no sense.
However this signature takes an array, not pointer:
int* test (int (&in)[5])
A variable referencing an array is basically a pointer to its first element, so yes, you can legitimately return a pointer to an array, because thery're essentially the same thing. Check this out yourself:
#include <assert.h>
int main() {
int a[] = {1, 2, 3, 4, 5};
int* pArr = a;
int* pFirstElem = &(a[0]);
assert(a == pArr);
assert(a == pFirstElem);
return 0;
}
This also means that passing an array to a function should be done via pointer (and not via int in[5]), and possibly along with the length of the array:
int* test(int* in, int len) {
int* out = in;
return out;
}
That said, you're right that using pointers (without fully understanding them) is pretty dangerous. For example, referencing an array that was allocated on the stack and went out of scope yields undefined behavior:
#include <iostream>
using namespace std;
int main() {
int* pArr = 0;
{
int a[] = {1, 2, 3, 4, 5};
pArr = a; // or test(a) if you wish
}
// a[] went out of scope here, but pArr holds a pointer to it
// all bets are off, this can output "1", output 1st chapter
// of "Romeo and Juliet", crash the program or destroy the
// universe
cout << pArr[0] << endl; // WRONG!
return 0;
}
So if you don't feel competent enough, just use std::vector.
[answer to the updated question]
The correct way to write your test function is either this:
void test(int* a, int* b, int* c, int len) {
for (int i = 0; i < len; ++i) c[i] = a[i] + b[i];
}
...
int main() {
int a[5] = {...}, b[5] = {...}, c[5] = {};
test(a, b, c, 5);
// c now holds the result
}
Or this (using std::vector):
#include <vector>
vector<int> test(const vector<int>& a, const vector<int>& b) {
vector<int> result(a.size());
for (int i = 0; i < a.size(); ++i) {
result[i] = a[i] + b[i];
}
return result; // copy will be elided
}
In a real app, the way you returned the array is called using an out parameter. Of course you don't actually have to return a pointer to the array, because the caller already has it, you just need to fill in the array. It's also common to pass another argument specifying the size of the array so as to not overflow it.
Using an out parameter has the disadvantage that the caller may not know how large the array needs to be to store the result. In that case, you can return a std::vector or similar array class instance.
Your code (which looks ok) doesn't return a pointer to an array. It returns a pointer to the first element of an array.
In fact that's usually what you want to do. Most manipulation of arrays are done via pointers to individual elements, not via pointers to the array as a whole.
You can define a pointer to an array, for example this:
double (*p)[42];
defines p as a pointer to a 42-element array of doubles. A big problem with that is that you have to specify the number of elements in the array as part of the type -- and that number has to be a compile-time constant. Most programs that deal with arrays need to deal with arrays of varying sizes; a given array's size won't vary after it's been created, but its initial size isn't necessarily known at compile time, and different array objects can have different sizes.
A pointer to the first element of an array lets you use either pointer arithmetic or the indexing operator [] to traverse the elements of the array. But the pointer doesn't tell you how many elements the array has; you generally have to keep track of that yourself.
If a function needs to create an array and return a pointer to its first element, you have to manage the storage for that array yourself, in one of several ways. You can have the caller pass in a pointer to (the first element of) an array object, probably along with another argument specifying its size -- which means the caller has to know how big the array needs to be. Or the function can return a pointer to (the first element of) a static array defined inside the function -- which means the size of the array is fixed, and the same array will be clobbered by a second call to the function. Or the function can allocate the array on the heap -- which makes the caller responsible for deallocating it later.
Everything I've written so far is common to C and C++, and in fact it's much more in the style of C than C++. Section 6 of the comp.lang.c FAQ discusses the behavior of arrays and pointers in C.
But if you're writing in C++, you're probably better off using C++ idioms. For example, the C++ standard library provides a number of headers defining container classes such as <vector> and <array>, which will take care of most of this stuff for you. Unless you have a particular reason to use raw arrays and pointers, you're probably better off just using C++ containers instead.
EDIT : I think you edited your question as I was typing this answer. The new code at the end of your question is, as you observer, no good; it returns a pointer to an object that ceases to exist as soon as the function returns. I think I've covered the alternatives.
you can (sort of) return an array
instead of
int m1[5] = {1, 2, 3, 4, 5};
int m2[5] = {6, 7, 8, 9, 10};
int* m3 = test(m1, m2);
write
struct mystruct
{
int arr[5];
};
int m1[5] = {1, 2, 3, 4, 5};
int m2[5] = {6, 7, 8, 9, 10};
mystruct m3 = test(m1,m2);
where test looks like
struct mystruct test(int m1[5], int m2[5])
{
struct mystruct s;
for (int i = 0; i < 5; ++i ) s.arr[i]=m1[i]+m2[i];
return s;
}
not very efficient since one is copying it delivers a copy of the array
I have an array int arr[5] that is passed to a function fillarr(int arr[]):
int fillarr(int arr[])
{
for(...);
return arr;
}
How can I return that array?
How will I use it, say I returned a pointer how am I going to access it?
In this case, your array variable arr can actually also be treated as a pointer to the beginning of your array's block in memory, by an implicit conversion. This syntax that you're using:
int fillarr(int arr[])
Is kind of just syntactic sugar. You could really replace it with this and it would still work:
int fillarr(int* arr)
So in the same sense, what you want to return from your function is actually a pointer to the first element in the array:
int* fillarr(int arr[])
And you'll still be able to use it just like you would a normal array:
int main()
{
int y[10];
int *a = fillarr(y);
cout << a[0] << endl;
}
C++ functions can't return C-style arrays by value. The closest thing is to return a pointer. Furthermore, an array type in the argument list is simply converted to a pointer.
int *fillarr( int arr[] ) { // arr "decays" to type int *
return arr;
}
You can improve it by using an array references for the argument and return, which prevents the decay:
int ( &fillarr( int (&arr)[5] ) )[5] { // no decay; argument must be size 5
return arr;
}
With Boost or C++11, pass-by-reference is only optional and the syntax is less mind-bending:
array< int, 5 > &fillarr( array< int, 5 > &arr ) {
return arr; // "array" being boost::array or std::array
}
The array template simply generates a struct containing a C-style array, so you can apply object-oriented semantics yet retain the array's original simplicity.
In C++11, you can return std::array.
#include <array>
using namespace std;
array<int, 5> fillarr(int arr[])
{
array<int, 5> arr2;
for(int i=0; i<5; ++i) {
arr2[i]=arr[i]*2;
}
return arr2;
}
$8.3.5/8 states-
"Functions shall not have a return type of type array or function, although they may have a return type of type pointer or reference to such things. There shall be no arrays of functions, although there can be arrays of pointers to functions."
int (&fn1(int (&arr)[5]))[5]{ // declare fn1 as returning refernce to array
return arr;
}
int *fn2(int arr[]){ // declare fn2 as returning pointer to array
return arr;
}
int main(){
int buf[5];
fn1(buf);
fn2(buf);
}
the answer may depend a bit on how you plan to use that function. For the simplest answer, lets decide that instead of an array, what you really want is a vector. Vectors are nice because the look for all the world like boring, ordinary values you can store in regular pointers. We'll look at other options and why you want them afterwards:
std::vector<int> fillarr( std::vector<int> arr ) {
// do something
return arr;
}
This will do exactly what you expect it to do. The upside is that std::vector takes care of making sure everything is handled cleanly. the downside is that this copies a very large amount of data, if your array is large. In fact it copies every element of the array twice. first it copies the vector so that the function can use it as a parameter. then it copies it again to return it to the caller. If you can handle managing the vector yourself, you can do things quite a bit more easily. (it may copy it a third time if the caller needs to store it in a variable of some sort to do more calculation)
It looks like what you're really trying to do is just populate a collection. if you don't have a specific reason to return a new instance of a collection, then don't. we can do it like this
void fillarr(std::vector<int> & arr) {
// modify arr
// don't return anything
}
this way you get a reference to the array passed to the function, not a private copy of it. any changes you make to the parameter are seen by the caller. You could return a reference to it if you want, but that's not really a great idea, since it sort of implies that you're getting something different from what you passed.
If you really do need a new instance of the collection, but want to avoid having it on the stack (and all the copying that entails), you need to create some kind of contract for how that instance is handled. the easiest way to do that is to use a smart pointer, which keeps the referenced instance around as long as anyone is holding onto it. It goes away cleanly if it goes out of scope. That would look like this.
std::auto_ptr<std::vector<int> > fillarr( const std::vector<int> & arr) {
std::auto_ptr<std::vector<int> > myArr(new std::vector<int>);
// do stuff with arr and *myArr
return myArr;
}
For the most part, using *myArr works identically to using a plain vanilla vector. This example also modifies the parameter list by adding the const keyword. Now you get a reference without copying it, but you can't modify it, so the caller knows it'll be the same as before the function got to it.
All of this is swell, but idiomatic c++ rarely works with collections as a whole. More normally, you will be using iterators over those collections. that would look something more like this
template <class Iterator>
Iterator fillarr(Iterator arrStart, Iterator arrEnd) {
Iterator arrIter = arrStart;
for(;arrIter <= arrEnd; arrIter++)
;// do something
return arrStart;
}
Using it looks a bit odd if you're not used to seeing this style.
vector<int> arr;
vector<int>::iterator foo = fillarr(arr.begin(), arr.end());
foo now 'points to' the beginning of the modified arr.
What's really nice about this is that it works equally well on vector as on plain C arrays and many other types of collection, for example
int arr[100];
int *foo = fillarr(arr, arr+100);
Which now looks an awful lot like the plain pointer examples given elsewhere in this question.
This:
int fillarr(int arr[])
is actually treated the same as:
int fillarr(int *arr)
Now if you really want to return an array you can change that line to
int * fillarr(int arr[]){
// do something to arr
return arr;
}
It's not really returning an array. you're returning a pointer to the start of the
array address.
But remember when you pass in the array, you're only passing in a pointer.
So when you modify the array data, you're actually modifying the data that the
pointer is pointing at. Therefore before you passed in the array, you must realise
that you already have on the outside the modified result.
e.g.
int fillarr(int arr[]){
array[0] = 10;
array[1] = 5;
}
int main(int argc, char* argv[]){
int arr[] = { 1,2,3,4,5 };
// arr[0] == 1
// arr[1] == 2 etc
int result = fillarr(arr);
// arr[0] == 10
// arr[1] == 5
return 0;
}
I suggest you might want to consider putting a length into your fillarr function like
this.
int * fillarr(int arr[], int length)
That way you can use length to fill the array to it's length no matter what it is.
To actually use it properly. Do something like this:
int * fillarr(int arr[], int length){
for (int i = 0; i < length; ++i){
// arr[i] = ? // do what you want to do here
}
return arr;
}
// then where you want to use it.
int arr[5];
int *arr2;
arr2 = fillarr(arr, 5);
// at this point, arr & arr2 are basically the same, just slightly
// different types. You can cast arr to a (char*) and it'll be the same.
If all you're wanting to do is set the array to some default values, consider using
the built in memset function.
something like:
memset((int*)&arr, 5, sizeof(int));
While I'm on the topic though. You say you're using C++. Have a look at using stl vectors. Your code is likely to be more robust.
There are lots of tutorials. Here is one that gives you an idea of how to use them.
http://www.yolinux.com/TUTORIALS/LinuxTutorialC++STL.html
This is a fairly old question, but I'm going to put in my 2 cents as there are a lot of answers, but none showing all possible methods in a clear and concise manner (not sure about the concise bit, as this got a bit out of hand. TL;DR 😉).
I'm assuming that the OP wanted to return the array that was passed in without copying as some means of directly passing this to the caller to be passed to another function to make the code look prettier.
However, to use an array like this is to let it decay into a pointer and have the compiler treat it like an array. This can result in subtle bugs if you pass in an array like, with the function expecting that it will have 5 elements, but your caller actually passes in some other number.
There a few ways you can handle this better. Pass in a std::vector or std::array (not sure if std::array was around in 2010 when the question was asked). You can then pass the object as a reference without any copying/moving of the object.
std::array<int, 5>& fillarr(std::array<int, 5>& arr)
{
// (before c++11)
for(auto it = arr.begin(); it != arr.end(); ++it)
{ /* do stuff */ }
// Note the following are for c++11 and higher. They will work for all
// the other examples below except for the stuff after the Edit.
// (c++11 and up)
for(auto it = std::begin(arr); it != std::end(arr); ++it)
{ /* do stuff */ }
// range for loop (c++11 and up)
for(auto& element : arr)
{ /* do stuff */ }
return arr;
}
std::vector<int>& fillarr(std::vector<int>& arr)
{
for(auto it = arr.begin(); it != arr.end(); ++it)
{ /* do stuff */ }
return arr;
}
However, if you insist on playing with C arrays, then use a template which will keep the information of how many items in the array.
template <size_t N>
int(&fillarr(int(&arr)[N]))[N]
{
// N is easier and cleaner than specifying sizeof(arr)/sizeof(arr[0])
for(int* it = arr; it != arr + N; ++it)
{ /* do stuff */ }
return arr;
}
Except, that looks butt ugly, and super hard to read. I now use something to help with that which wasn't around in 2010, which I also use for function pointers:
template <typename T>
using type_t = T;
template <size_t N>
type_t<int(&)[N]> fillarr(type_t<int(&)[N]> arr)
{
// N is easier and cleaner than specifying sizeof(arr)/sizeof(arr[0])
for(int* it = arr; it != arr + N; ++it)
{ /* do stuff */ }
return arr;
}
This moves the type where one would expect it to be, making this far more readable. Of course, using a template is superfluous if you are not going to use anything but 5 elements, so you can of course hard code it:
type_t<int(&)[5]> fillarr(type_t<int(&)[5]> arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
As I said, my type_t<> trick wouldn't have worked at the time this question was asked. The best you could have hoped for back then was to use a type in a struct:
template<typename T>
struct type
{
typedef T type;
};
typename type<int(&)[5]>::type fillarr(typename type<int(&)[5]>::type arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
Which starts to look pretty ugly again, but at least is still more readable, though the typename may have been optional back then depending on the compiler, resulting in:
type<int(&)[5]>::type fillarr(type<int(&)[5]>::type arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
And then of course you could have specified a specific type, rather than using my helper.
typedef int(&array5)[5];
array5 fillarr(array5 arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
Back then, the free functions std::begin() and std::end() didn't exist, though could have been easily implemented. This would have allowed iterating over the array in a safer manner as they make sense on a C array, but not a pointer.
As for accessing the array, you could either pass it to another function that takes the same parameter type, or make an alias to it (which wouldn't make much sense as you already have the original in that scope). Accessing a array reference is just like accessing the original array.
void other_function(type_t<int(&)[5]> x) { /* do something else */ }
void fn()
{
int array[5];
other_function(fillarr(array));
}
or
void fn()
{
int array[5];
auto& array2 = fillarr(array); // alias. But why bother.
int forth_entry = array[4];
int forth_entry2 = array2[4]; // same value as forth_entry
}
To summarize, it is best to not allow an array decay into a pointer if you intend to iterate over it. It is just a bad idea as it keeps the compiler from protecting you from shooting yourself in the foot and makes your code harder to read. Always try and help the compiler help you by keeping the types as long as possible unless you have a very good reason not to do so.
Edit
Oh, and for completeness, you can allow it to degrade to a pointer, but this decouples the array from the number of elements it holds. This is done a lot in C/C++ and is usually mitigated by passing the number of elements in the array. However, the compiler can't help you if you make a mistake and pass in the wrong value to the number of elements.
// separate size value
int* fillarr(int* arr, size_t size)
{
for(int* it = arr; it != arr + size; ++it)
{ /* do stuff */ }
return arr;
}
Instead of passing the size, you can pass the end pointer, which will point to one past the end of your array. This is useful as it makes for something that is closer to the std algorithms, which take a begin and and end pointer, but what you return is now only something that you must remember.
// separate end pointer
int* fillarr(int* arr, int* end)
{
for(int* it = arr; it != end; ++it)
{ /* do stuff */ }
return arr;
}
Alternatively, you can document that this function will only take 5 elements and hope that the user of your function doesn't do anything stupid.
// I document that this function will ONLY take 5 elements and
// return the same array of 5 elements. If you pass in anything
// else, may nazal demons exit thine nose!
int* fillarr(int* arr)
{
for(int* it = arr; it != arr + 5; ++it)
{ /* do stuff */ }
return arr;
}
Note that the return value has lost it's original type and is degraded to a pointer. Because of this, you are now on your own to ensure that you are not going to overrun the array.
You could pass a std::pair<int*, int*>, which you can use for begin and end and pass that around, but then it really stops looking like an array.
std::pair<int*, int*> fillarr(std::pair<int*, int*> arr)
{
for(int* it = arr.first; it != arr.second; ++it)
{ /* do stuff */ }
return arr; // if you change arr, then return the original arr value.
}
void fn()
{
int array[5];
auto array2 = fillarr(std::make_pair(&array[0], &array[5]));
// Can be done, but you have the original array in scope, so why bother.
int fourth_element = array2.first[4];
}
or
void other_function(std::pair<int*, int*> array)
{
// Can be done, but you have the original array in scope, so why bother.
int fourth_element = array2.first[4];
}
void fn()
{
int array[5];
other_function(fillarr(std::make_pair(&array[0], &array[5])));
}
Funny enough, this is very similar to how std::initializer_list work (c++11), but they don't work in this context.
to return an array from a function , let us define that array in a structure;
So it looks something like this
struct Marks{
int list[5];
}
Now let us create variables of the type structure.
typedef struct Marks marks;
marks marks_list;
We can pass array to a function in the following way and assign value to it:
void setMarks(int marks_array[]){
for(int i=0;i<sizeof(marks_array)/sizeof(int);i++)
marks_list.list[i]=marks_array[i];
}
We can also return the array. To return the array , the return type of the function should be of structure type ie marks. This is because in reality we are passing the structure that contains the array. So the final code may look like this.
marks getMarks(){
return marks_list;
}
the Simplest way to do this ,is to return it by reference , even if you don't write
the '&' symbol , it is automatically returned by reference
void fillarr(int arr[5])
{
for(...);
}
int *fillarr(int arr[])
You can still use the result like
int *returned_array = fillarr(some_other_array);
if(returned_array[0] == 3)
do_important_cool_stuff();
As above mentioned paths are correct. But i think if we just return a local array variable of a function sometimes it returns garbage values as its elements.
in-order to avoid that i had to create the array dynamically and proceed. Which is something like this.
int* func()
{
int* Arr = new int[100];
return Arr;
}
int main()
{
int* ArrResult = func();
cout << ArrResult[0] << " " << ArrResult[1] << endl;
return 0;
}
Source: https://www.tutorialspoint.com/cplusplus/cpp_return_arrays_from_functions.htm
C++ does not allow to return an entire array as an argument to a function. However, you can return a pointer to an array by specifying the array's name without an index.
If you want to return a single-dimension array from a function, you would have to declare a function returning a pointer as in the following example:
int * myFunction() {
.
.
.
}
C++ does not advocate to return the address of a local variable to outside of the function so you would have to define the local variable as static variable.
Applying these rules on the current question, we can write the program as follows:
# include <iostream>
using namespace std;
int * fillarr( );
int main ()
{
int *p;
p = fillarr();
for ( int i = 0; i < 5; i++ )
cout << "p[" << i << "] : "<< *(p + i) << endl;
return 0;
}
int * fillarr( )
{
static int arr[5];
for (int i = 0; i < 5; ++i)
arr[i] = i;
return arr;
}
The Output will be:
p[0]=0
p[1]=1
p[2]=2
p[3]=3
p[4]=4
template<typename T, size_t N>
using ARR_REF = T (&)[N];
template <typename T, size_t N>
ARR_REF<T,N> ArraySizeHelper(ARR_REF<T,N> arr);
#define arraysize(arr) sizeof(ArraySizeHelper(arr))
and what about:
int (*func())
{
int *f = new int[10] {1,2,3};
return f;
}
int fa[10] = { 0 };
auto func2() -> int (*) [10]
{
return &fa;
}
Actually when you pass an array inside a function, the pointer to the original array is passed in the function parameter and thus the changes made to the array inside that function is actually made on the original array.
#include <iostream>
using namespace std;
int* func(int ar[])
{
for(int i=0;i<100;i++)
ar[i]=i;
int *ptr=ar;
return ptr;
}
int main() {
int *p;
int y[100]={0};
p=func(y);
for(int i=0;i<100;i++)
cout<<i<<" : "<<y[i]<<'\n';
}
Run it and you will see the changes
And why don't "return" the array as a parameter?
fillarr(int source[], size_t dimSource, int dest[], size_t dimDest)
{
if (dimSource <= dimDest)
{
for (size_t i = 0; i < dimSource; i++)
{
//some stuff...
}
}
else
{
//some stuff..
}
}
or..in a simpler way (but you have to know the dimensions...):
fillarr(int source[], int dest[])
{
//...
}
A simple and elaborate example, so that I can refer here if I forget the concept and need help.
#include <iostream>
using namespace std;
int *ReturnArray(int arr[], int size)
{
static int MinMax[2] = {0, 0}; // must use static, else address would be deleted after the return is reached
MinMax[0] = arr[0];
MinMax[1] = arr[size - 1];
return MinMax;
}
int main()
{
int arr[] = {1, 2, 3};
int size = sizeof(arr) / sizeof(*arr);
int *ans; // pointer to hold returned array
ans = ReturnArray(arr, size); // only pointer can receive the return, not an array
cout << "Min: " << ans[0] << " Max: " << ans[1];
return 0;
}
Here's a full example of this kind of problem to solve
#include <bits/stdc++.h>
using namespace std;
int* solve(int brr[],int n)
{
sort(brr,brr+n);
return brr;
}
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
int *a=solve(arr,n);
for(int i=0;i<n;i++)
{
cout<<a[i]<<endl;
}
return 0;
}
i used static array so that while returning array it should not throw error as you are returning address of local variable...
so now you can send any locally created variable from function by making it as static...as it works as global variable....
#include<iostream>
using namespace std;
char *func(int n)
{
// char a[26]; /*if we use this then an error will occur because you are
// returning address of a local variable*/
static char a[26];
char temp='A';
for(int i=0;i<n;i++)
{
a[i]=temp;temp++;
}
return a;
}
int main()
{
int n=26;
char *p=func(n);
for(int i=0;i<n;i++)
cout<<*(p+i)<<" ";
//or you can also print like this
for(int i=0;i<n;i++)
cout<<p[i]<<" ";
}
Just define a type[ ] as return value, like:
private string[] functionReturnValueArray(string one, string two)
{
string[] x = {one, two};
x[0] = "a";
x[1] = "b";
return x;
}
.
.
.
function call:
string[] y;
y = functionReturnValueArray(stringOne, stringTwo)
I got this problem which I could not explain today.
int main(){
vector<int> vec = subroutine();
...
delete(&vec);
}
vector<int>& subroutine(){
vector<int>* vec = new vector<int>();
//Init the vec
for(int i=0;i<vec.size();i++){
...
}
return *vec;
}
And it came with an error as:
double free or corruption (out): 0X00007fffa145ff50
It's definitely the problem with the line:
delete(&vec);
But I just can't explain, why there is a double free?
You are calling delete on something that is not allocated with new because you are making a copy of the newed vector. You could "correct" your code by not copying the vector:
vector<int>& vec = subroutine();
but this kind of function really is asking for trouble. The correct way to fix this would be to return a vector by value.
vector<int> subroutine() {
vector<int> vec;
//Init the vec
for(int i=0;i<vec.size();i++){
...
}
return vec;
}
When you return from the function, a copy of the vector is made and, at this point, the newed memory is leaked and cannot be reached; hence the error, you are trying to delete something on the stack (and not on the heap).
I don't see a reason why you need the new; you could try this instead.
vector<int> subroutine(){
vector<int> vec;
//Init the vec
for(int i=0;i<vec.size();i++){
//...
}
return vec;
}
Return the vector by value.
vector<int> vec = subroutine(); here you make a copy. You do not return the pointer. You return a whole object that the pointer refers to, on this line return *vec;.
Thus, the vec from main is deleted explicitly by you, and then again when its scope ends.
Moreover, the vec from subroutine is never freed.
I'm trying to return a pointer to an array from a function but I have an issue. When I try to output like this:
#include <iostream>
using namespace std;
int* Somma_Array(int[],int[],int);
int main()
{
int n;
cin>>n;
int A[n],B[n];
for(int i=0;i<n;i++)cin>>A[i];
for(int i=0;i<n;i++)cin>>B[i];
int *c=Somma_Array(A,B,n);
for(int i=0;i<n*2;i++)cout<<c[i];
}
int* Somma_Array(int v[],int p[],int size)
{
int r[size*2];
for(int i=0;i<size;i++)r[i]=v[i];
for(int i=0;i<size;i++)r[i+size]=p[i];
return r;
}
it prints weird numbers instead of the actual number. I tried to do what this question says but it does not work. It gives me the following warning:
[Warning] address of local variable 'r' returned [enabled by default]
I'm using bloodshed dev-c++.
You define a stack allocated array r, which is destroyed when you exit the function Soma_Array. This is one of the (many) reasons vectors are preferred to plain arrays - they handle allocation and deallocation for you.
#include <vector>
std::vector<int> getArray()
{
std::vector<int> a = {1, 2, 3};
return a;
}
The following:
int r[size*2];
defines r locally. When the function exits (as in the scope of the function expires), r will be destroyed since it is bound to the function's scope. You are likely seeing junk data from the stack frame.
you could fix this by doing the following:
int* r = new int[size * 2];
The variable r will now be heap allocated and exist beyond the scope of the function.
IMPORTANT by doing this, you now must manually free r when you are done with it. So for instance, your calling code will look something like this:
int* result = Somma_Array(v, p, size);
/* ... do stuff ... */
delete[] result;
Since r is an array, note the use of delete[] instead of delete. delete[] is the correct way to destroy arrays.
A Better Alternative
Would std::vector be more what you are after? This is a much safer alternative to hand-rolled arrays. The vector is safer to use, scales automatically as you add elements, and cleans itself up nicely when it leaves scope (assuming you are using a value-type instance). Additionally, vectors can be copied and moved out of functions easily.
You cannot return arrays in C++. Especially, you should not return a pointer to a local array. You can however return a std::vector<int>:
std::vector<int> Somma_Array(int v[], int p[], int size)
{
std::vector<int> r(2 * size);
std::copy(v, v + size, r.begin());
std::copy(p, p + size, r.begin() + size);
return r;
}
I have an array int arr[5] that is passed to a function fillarr(int arr[]):
int fillarr(int arr[])
{
for(...);
return arr;
}
How can I return that array?
How will I use it, say I returned a pointer how am I going to access it?
In this case, your array variable arr can actually also be treated as a pointer to the beginning of your array's block in memory, by an implicit conversion. This syntax that you're using:
int fillarr(int arr[])
Is kind of just syntactic sugar. You could really replace it with this and it would still work:
int fillarr(int* arr)
So in the same sense, what you want to return from your function is actually a pointer to the first element in the array:
int* fillarr(int arr[])
And you'll still be able to use it just like you would a normal array:
int main()
{
int y[10];
int *a = fillarr(y);
cout << a[0] << endl;
}
C++ functions can't return C-style arrays by value. The closest thing is to return a pointer. Furthermore, an array type in the argument list is simply converted to a pointer.
int *fillarr( int arr[] ) { // arr "decays" to type int *
return arr;
}
You can improve it by using an array references for the argument and return, which prevents the decay:
int ( &fillarr( int (&arr)[5] ) )[5] { // no decay; argument must be size 5
return arr;
}
With Boost or C++11, pass-by-reference is only optional and the syntax is less mind-bending:
array< int, 5 > &fillarr( array< int, 5 > &arr ) {
return arr; // "array" being boost::array or std::array
}
The array template simply generates a struct containing a C-style array, so you can apply object-oriented semantics yet retain the array's original simplicity.
In C++11, you can return std::array.
#include <array>
using namespace std;
array<int, 5> fillarr(int arr[])
{
array<int, 5> arr2;
for(int i=0; i<5; ++i) {
arr2[i]=arr[i]*2;
}
return arr2;
}
$8.3.5/8 states-
"Functions shall not have a return type of type array or function, although they may have a return type of type pointer or reference to such things. There shall be no arrays of functions, although there can be arrays of pointers to functions."
int (&fn1(int (&arr)[5]))[5]{ // declare fn1 as returning refernce to array
return arr;
}
int *fn2(int arr[]){ // declare fn2 as returning pointer to array
return arr;
}
int main(){
int buf[5];
fn1(buf);
fn2(buf);
}
the answer may depend a bit on how you plan to use that function. For the simplest answer, lets decide that instead of an array, what you really want is a vector. Vectors are nice because the look for all the world like boring, ordinary values you can store in regular pointers. We'll look at other options and why you want them afterwards:
std::vector<int> fillarr( std::vector<int> arr ) {
// do something
return arr;
}
This will do exactly what you expect it to do. The upside is that std::vector takes care of making sure everything is handled cleanly. the downside is that this copies a very large amount of data, if your array is large. In fact it copies every element of the array twice. first it copies the vector so that the function can use it as a parameter. then it copies it again to return it to the caller. If you can handle managing the vector yourself, you can do things quite a bit more easily. (it may copy it a third time if the caller needs to store it in a variable of some sort to do more calculation)
It looks like what you're really trying to do is just populate a collection. if you don't have a specific reason to return a new instance of a collection, then don't. we can do it like this
void fillarr(std::vector<int> & arr) {
// modify arr
// don't return anything
}
this way you get a reference to the array passed to the function, not a private copy of it. any changes you make to the parameter are seen by the caller. You could return a reference to it if you want, but that's not really a great idea, since it sort of implies that you're getting something different from what you passed.
If you really do need a new instance of the collection, but want to avoid having it on the stack (and all the copying that entails), you need to create some kind of contract for how that instance is handled. the easiest way to do that is to use a smart pointer, which keeps the referenced instance around as long as anyone is holding onto it. It goes away cleanly if it goes out of scope. That would look like this.
std::auto_ptr<std::vector<int> > fillarr( const std::vector<int> & arr) {
std::auto_ptr<std::vector<int> > myArr(new std::vector<int>);
// do stuff with arr and *myArr
return myArr;
}
For the most part, using *myArr works identically to using a plain vanilla vector. This example also modifies the parameter list by adding the const keyword. Now you get a reference without copying it, but you can't modify it, so the caller knows it'll be the same as before the function got to it.
All of this is swell, but idiomatic c++ rarely works with collections as a whole. More normally, you will be using iterators over those collections. that would look something more like this
template <class Iterator>
Iterator fillarr(Iterator arrStart, Iterator arrEnd) {
Iterator arrIter = arrStart;
for(;arrIter <= arrEnd; arrIter++)
;// do something
return arrStart;
}
Using it looks a bit odd if you're not used to seeing this style.
vector<int> arr;
vector<int>::iterator foo = fillarr(arr.begin(), arr.end());
foo now 'points to' the beginning of the modified arr.
What's really nice about this is that it works equally well on vector as on plain C arrays and many other types of collection, for example
int arr[100];
int *foo = fillarr(arr, arr+100);
Which now looks an awful lot like the plain pointer examples given elsewhere in this question.
This:
int fillarr(int arr[])
is actually treated the same as:
int fillarr(int *arr)
Now if you really want to return an array you can change that line to
int * fillarr(int arr[]){
// do something to arr
return arr;
}
It's not really returning an array. you're returning a pointer to the start of the
array address.
But remember when you pass in the array, you're only passing in a pointer.
So when you modify the array data, you're actually modifying the data that the
pointer is pointing at. Therefore before you passed in the array, you must realise
that you already have on the outside the modified result.
e.g.
int fillarr(int arr[]){
array[0] = 10;
array[1] = 5;
}
int main(int argc, char* argv[]){
int arr[] = { 1,2,3,4,5 };
// arr[0] == 1
// arr[1] == 2 etc
int result = fillarr(arr);
// arr[0] == 10
// arr[1] == 5
return 0;
}
I suggest you might want to consider putting a length into your fillarr function like
this.
int * fillarr(int arr[], int length)
That way you can use length to fill the array to it's length no matter what it is.
To actually use it properly. Do something like this:
int * fillarr(int arr[], int length){
for (int i = 0; i < length; ++i){
// arr[i] = ? // do what you want to do here
}
return arr;
}
// then where you want to use it.
int arr[5];
int *arr2;
arr2 = fillarr(arr, 5);
// at this point, arr & arr2 are basically the same, just slightly
// different types. You can cast arr to a (char*) and it'll be the same.
If all you're wanting to do is set the array to some default values, consider using
the built in memset function.
something like:
memset((int*)&arr, 5, sizeof(int));
While I'm on the topic though. You say you're using C++. Have a look at using stl vectors. Your code is likely to be more robust.
There are lots of tutorials. Here is one that gives you an idea of how to use them.
http://www.yolinux.com/TUTORIALS/LinuxTutorialC++STL.html
This is a fairly old question, but I'm going to put in my 2 cents as there are a lot of answers, but none showing all possible methods in a clear and concise manner (not sure about the concise bit, as this got a bit out of hand. TL;DR 😉).
I'm assuming that the OP wanted to return the array that was passed in without copying as some means of directly passing this to the caller to be passed to another function to make the code look prettier.
However, to use an array like this is to let it decay into a pointer and have the compiler treat it like an array. This can result in subtle bugs if you pass in an array like, with the function expecting that it will have 5 elements, but your caller actually passes in some other number.
There a few ways you can handle this better. Pass in a std::vector or std::array (not sure if std::array was around in 2010 when the question was asked). You can then pass the object as a reference without any copying/moving of the object.
std::array<int, 5>& fillarr(std::array<int, 5>& arr)
{
// (before c++11)
for(auto it = arr.begin(); it != arr.end(); ++it)
{ /* do stuff */ }
// Note the following are for c++11 and higher. They will work for all
// the other examples below except for the stuff after the Edit.
// (c++11 and up)
for(auto it = std::begin(arr); it != std::end(arr); ++it)
{ /* do stuff */ }
// range for loop (c++11 and up)
for(auto& element : arr)
{ /* do stuff */ }
return arr;
}
std::vector<int>& fillarr(std::vector<int>& arr)
{
for(auto it = arr.begin(); it != arr.end(); ++it)
{ /* do stuff */ }
return arr;
}
However, if you insist on playing with C arrays, then use a template which will keep the information of how many items in the array.
template <size_t N>
int(&fillarr(int(&arr)[N]))[N]
{
// N is easier and cleaner than specifying sizeof(arr)/sizeof(arr[0])
for(int* it = arr; it != arr + N; ++it)
{ /* do stuff */ }
return arr;
}
Except, that looks butt ugly, and super hard to read. I now use something to help with that which wasn't around in 2010, which I also use for function pointers:
template <typename T>
using type_t = T;
template <size_t N>
type_t<int(&)[N]> fillarr(type_t<int(&)[N]> arr)
{
// N is easier and cleaner than specifying sizeof(arr)/sizeof(arr[0])
for(int* it = arr; it != arr + N; ++it)
{ /* do stuff */ }
return arr;
}
This moves the type where one would expect it to be, making this far more readable. Of course, using a template is superfluous if you are not going to use anything but 5 elements, so you can of course hard code it:
type_t<int(&)[5]> fillarr(type_t<int(&)[5]> arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
As I said, my type_t<> trick wouldn't have worked at the time this question was asked. The best you could have hoped for back then was to use a type in a struct:
template<typename T>
struct type
{
typedef T type;
};
typename type<int(&)[5]>::type fillarr(typename type<int(&)[5]>::type arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
Which starts to look pretty ugly again, but at least is still more readable, though the typename may have been optional back then depending on the compiler, resulting in:
type<int(&)[5]>::type fillarr(type<int(&)[5]>::type arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
And then of course you could have specified a specific type, rather than using my helper.
typedef int(&array5)[5];
array5 fillarr(array5 arr)
{
// Prefer using the compiler to figure out how many elements there are
// as it reduces the number of locations where you have to change if needed.
for(int* it = arr; it != arr + sizeof(arr)/sizeof(arr[0]); ++it)
{ /* do stuff */ }
return arr;
}
Back then, the free functions std::begin() and std::end() didn't exist, though could have been easily implemented. This would have allowed iterating over the array in a safer manner as they make sense on a C array, but not a pointer.
As for accessing the array, you could either pass it to another function that takes the same parameter type, or make an alias to it (which wouldn't make much sense as you already have the original in that scope). Accessing a array reference is just like accessing the original array.
void other_function(type_t<int(&)[5]> x) { /* do something else */ }
void fn()
{
int array[5];
other_function(fillarr(array));
}
or
void fn()
{
int array[5];
auto& array2 = fillarr(array); // alias. But why bother.
int forth_entry = array[4];
int forth_entry2 = array2[4]; // same value as forth_entry
}
To summarize, it is best to not allow an array decay into a pointer if you intend to iterate over it. It is just a bad idea as it keeps the compiler from protecting you from shooting yourself in the foot and makes your code harder to read. Always try and help the compiler help you by keeping the types as long as possible unless you have a very good reason not to do so.
Edit
Oh, and for completeness, you can allow it to degrade to a pointer, but this decouples the array from the number of elements it holds. This is done a lot in C/C++ and is usually mitigated by passing the number of elements in the array. However, the compiler can't help you if you make a mistake and pass in the wrong value to the number of elements.
// separate size value
int* fillarr(int* arr, size_t size)
{
for(int* it = arr; it != arr + size; ++it)
{ /* do stuff */ }
return arr;
}
Instead of passing the size, you can pass the end pointer, which will point to one past the end of your array. This is useful as it makes for something that is closer to the std algorithms, which take a begin and and end pointer, but what you return is now only something that you must remember.
// separate end pointer
int* fillarr(int* arr, int* end)
{
for(int* it = arr; it != end; ++it)
{ /* do stuff */ }
return arr;
}
Alternatively, you can document that this function will only take 5 elements and hope that the user of your function doesn't do anything stupid.
// I document that this function will ONLY take 5 elements and
// return the same array of 5 elements. If you pass in anything
// else, may nazal demons exit thine nose!
int* fillarr(int* arr)
{
for(int* it = arr; it != arr + 5; ++it)
{ /* do stuff */ }
return arr;
}
Note that the return value has lost it's original type and is degraded to a pointer. Because of this, you are now on your own to ensure that you are not going to overrun the array.
You could pass a std::pair<int*, int*>, which you can use for begin and end and pass that around, but then it really stops looking like an array.
std::pair<int*, int*> fillarr(std::pair<int*, int*> arr)
{
for(int* it = arr.first; it != arr.second; ++it)
{ /* do stuff */ }
return arr; // if you change arr, then return the original arr value.
}
void fn()
{
int array[5];
auto array2 = fillarr(std::make_pair(&array[0], &array[5]));
// Can be done, but you have the original array in scope, so why bother.
int fourth_element = array2.first[4];
}
or
void other_function(std::pair<int*, int*> array)
{
// Can be done, but you have the original array in scope, so why bother.
int fourth_element = array2.first[4];
}
void fn()
{
int array[5];
other_function(fillarr(std::make_pair(&array[0], &array[5])));
}
Funny enough, this is very similar to how std::initializer_list work (c++11), but they don't work in this context.
to return an array from a function , let us define that array in a structure;
So it looks something like this
struct Marks{
int list[5];
}
Now let us create variables of the type structure.
typedef struct Marks marks;
marks marks_list;
We can pass array to a function in the following way and assign value to it:
void setMarks(int marks_array[]){
for(int i=0;i<sizeof(marks_array)/sizeof(int);i++)
marks_list.list[i]=marks_array[i];
}
We can also return the array. To return the array , the return type of the function should be of structure type ie marks. This is because in reality we are passing the structure that contains the array. So the final code may look like this.
marks getMarks(){
return marks_list;
}
the Simplest way to do this ,is to return it by reference , even if you don't write
the '&' symbol , it is automatically returned by reference
void fillarr(int arr[5])
{
for(...);
}
int *fillarr(int arr[])
You can still use the result like
int *returned_array = fillarr(some_other_array);
if(returned_array[0] == 3)
do_important_cool_stuff();
As above mentioned paths are correct. But i think if we just return a local array variable of a function sometimes it returns garbage values as its elements.
in-order to avoid that i had to create the array dynamically and proceed. Which is something like this.
int* func()
{
int* Arr = new int[100];
return Arr;
}
int main()
{
int* ArrResult = func();
cout << ArrResult[0] << " " << ArrResult[1] << endl;
return 0;
}
Source: https://www.tutorialspoint.com/cplusplus/cpp_return_arrays_from_functions.htm
C++ does not allow to return an entire array as an argument to a function. However, you can return a pointer to an array by specifying the array's name without an index.
If you want to return a single-dimension array from a function, you would have to declare a function returning a pointer as in the following example:
int * myFunction() {
.
.
.
}
C++ does not advocate to return the address of a local variable to outside of the function so you would have to define the local variable as static variable.
Applying these rules on the current question, we can write the program as follows:
# include <iostream>
using namespace std;
int * fillarr( );
int main ()
{
int *p;
p = fillarr();
for ( int i = 0; i < 5; i++ )
cout << "p[" << i << "] : "<< *(p + i) << endl;
return 0;
}
int * fillarr( )
{
static int arr[5];
for (int i = 0; i < 5; ++i)
arr[i] = i;
return arr;
}
The Output will be:
p[0]=0
p[1]=1
p[2]=2
p[3]=3
p[4]=4
template<typename T, size_t N>
using ARR_REF = T (&)[N];
template <typename T, size_t N>
ARR_REF<T,N> ArraySizeHelper(ARR_REF<T,N> arr);
#define arraysize(arr) sizeof(ArraySizeHelper(arr))
and what about:
int (*func())
{
int *f = new int[10] {1,2,3};
return f;
}
int fa[10] = { 0 };
auto func2() -> int (*) [10]
{
return &fa;
}
Actually when you pass an array inside a function, the pointer to the original array is passed in the function parameter and thus the changes made to the array inside that function is actually made on the original array.
#include <iostream>
using namespace std;
int* func(int ar[])
{
for(int i=0;i<100;i++)
ar[i]=i;
int *ptr=ar;
return ptr;
}
int main() {
int *p;
int y[100]={0};
p=func(y);
for(int i=0;i<100;i++)
cout<<i<<" : "<<y[i]<<'\n';
}
Run it and you will see the changes
And why don't "return" the array as a parameter?
fillarr(int source[], size_t dimSource, int dest[], size_t dimDest)
{
if (dimSource <= dimDest)
{
for (size_t i = 0; i < dimSource; i++)
{
//some stuff...
}
}
else
{
//some stuff..
}
}
or..in a simpler way (but you have to know the dimensions...):
fillarr(int source[], int dest[])
{
//...
}
A simple and elaborate example, so that I can refer here if I forget the concept and need help.
#include <iostream>
using namespace std;
int *ReturnArray(int arr[], int size)
{
static int MinMax[2] = {0, 0}; // must use static, else address would be deleted after the return is reached
MinMax[0] = arr[0];
MinMax[1] = arr[size - 1];
return MinMax;
}
int main()
{
int arr[] = {1, 2, 3};
int size = sizeof(arr) / sizeof(*arr);
int *ans; // pointer to hold returned array
ans = ReturnArray(arr, size); // only pointer can receive the return, not an array
cout << "Min: " << ans[0] << " Max: " << ans[1];
return 0;
}
Here's a full example of this kind of problem to solve
#include <bits/stdc++.h>
using namespace std;
int* solve(int brr[],int n)
{
sort(brr,brr+n);
return brr;
}
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
int *a=solve(arr,n);
for(int i=0;i<n;i++)
{
cout<<a[i]<<endl;
}
return 0;
}
i used static array so that while returning array it should not throw error as you are returning address of local variable...
so now you can send any locally created variable from function by making it as static...as it works as global variable....
#include<iostream>
using namespace std;
char *func(int n)
{
// char a[26]; /*if we use this then an error will occur because you are
// returning address of a local variable*/
static char a[26];
char temp='A';
for(int i=0;i<n;i++)
{
a[i]=temp;temp++;
}
return a;
}
int main()
{
int n=26;
char *p=func(n);
for(int i=0;i<n;i++)
cout<<*(p+i)<<" ";
//or you can also print like this
for(int i=0;i<n;i++)
cout<<p[i]<<" ";
}
Just define a type[ ] as return value, like:
private string[] functionReturnValueArray(string one, string two)
{
string[] x = {one, two};
x[0] = "a";
x[1] = "b";
return x;
}
.
.
.
function call:
string[] y;
y = functionReturnValueArray(stringOne, stringTwo)