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I would like to create a struct which contains both an int and an array of int. So I define it like
struct my_struct {
int N ;
int arr[30] ;
int arr[30][30] ;
}
Then I would like to initialize it with an array which I have already defined and initialized, for example
int my_arr[30] ;
for (int i = 0; i < 30; ++i)
{
my_arr[i] = i ;
}
Then I thought I could initialize a struct as
my_struct A = {30,my_arr}
but it doesn't seem to work (gives conversion error).
P.s. and how would it work with a 2d array?
Arrays cannot be copy-initialised. This isn't particular to the array being member of a class; same can be reproduced like this:
int a[30] = {};
int b[30] = a; // ill-formed
You can initialise array elements like this:
my_struct A = {30, {my_arr[0], my_arr[1], my_arr[2], //...
But that's not always very convenient. Alternatively, you can assign the array values using a loop like you did initially. You don't have to write the loop either, there's a standard algorithm for this called std::copy.
#include <iostream>
#include <array>
using namespace std;
int main(){
int a = [];
int b = 10;
std::fill(a);
cout<<a<<endl;
}
I have an array "a" and want to fill it with an integer "b". As I remember in python its simply uses apppend, does someone know solution?
Here one solution how to use your array header.
int b = 10;
std::array<int, 3> a;
std::fill(begin(a), end(a), b);
I have an array "a"
int a = [];
What you have is a syntax error.
As I remember in python its simply uses apppend
A major difference between a C++ array and python list is that the size of C++ array cannot change, and thus nothing can be appended into it.
How to fill array in C++?
There is indeed a standard algorithm for this purpose:
int a[4];
int b = 10;
std::fill_n(a, std::size(a), b);
Decide the size for a, as it is an array not a list. For example:
int a[10];
Then use index values to fill in the array like:
a[0] = 1;
a[1] = 4;
etc.
If you want a dynamic array use std::vector instead
Here is how its done with vectors:
std::vector<int> myvector;
int myint = 3;
myvector.push_back (myint);
Following zohaib's answer:
If your array is of fixed length:
You can use the array's 'fill' member function like this:
a.fill(b);
If your array can change it's size you can use the std's fill function like this:
std::fill(a.begin(), a.end(), b);
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)
It is possible to give an initializer list to the definition of a static array. Example:
int main()
{
int int_static[2] = {1,2};
}
Is a similar initializer list possible for a dynamic array?
int main()
{
int* int_ptr = new int[2];
}
This is closer to what I am trying to do:
struct foo
{
foo(){}
foo(void * ptr): ptr_(ptr) {}
void * ptr_;
};
int main()
{
foo* foo_ptr = new foo[10];
}
At initialization time not the default constructor should be called, but foo:foo(void*).
The point of having a static initializer list for a dynamic array might come handy in the case of Just-In-Time compilation for accelerator cores which do have only a limited amount of stack available, but at the same time you construct your objects with a (accelerator compile time = host run time) static initializer list.
I assume not (since this would require the compiler to generate additional code, namely to copy the values of the arguments to the heap location). I think c++0x supports some of this, but I cannot use it.
Right now I could use such a construct. Maybe someone knows a trick..
Best!
At the time the OP posted this question, C++11 support may not have been very prevalent yet, which is why the accepted answer says this is not possible. However, initializing a dynamic array with an explicit initializer list should now be supported in all major C++ compilers.
The syntax new int[3] {1, 2, 3} was standardized in C++11. Quoting the new expression page on cppreference.com:
The object created by a new-expression is initialized according to the following rules:
...
If type is an array type, an array of objects is initialized:
...
If initializer is a brace-enclosed list of arguments, the array is aggregate-initialized. (since C++11)
So, given the OP's example, the following is perfectly legal when using C++11 or newer:
foo * foo_array = new foo[2] { nullptr, nullptr };
Note that by providing pointers in the initializer list, we're actually coaxing the compiler to apply the foo(void * ptr) constructor (rather than the default constructor), which was the desired behavior.
No, you cannot do that.
I think C++ doesn't allow this because allowing such thing doesn't add any nice-to-have feature to the language. In other words, what would be the point of dynamic array if you use a static initializer to initialize it?
The point of dynamic array is to create an array of size N which is known at runtime, depending on the actual need. That is, the code
int *p = new int[2];
makes less sense to me than the following:
int *p = new int[N]; //N is known at runtime
If that is so, then how can you provide the number of elements in the static initializer because N isn't known until runtime?
Lets assume that you're allowed to write this:
int *p = new int[2] {10,20}; //pretend this!
But what big advantage are you getting by writing this? Nothing. Its almost same as:
int a[] = {10,20};
The real advantage would be when you're allowed to write that for arrays of N elements. But then the problem is this:
int *p = new int[N] {10,20, ... /*Oops, no idea how far we can go? N is not known!*/ };
No, you will have to create the elements dynamically.
Alternatively, you can use a local array and copy its elements over those of the dynamically allocated array:
int main() {
int _detail[] = { 1, 2 };
int * ptr = new int[2];
std::copy( _detail, _detail+(sizeof detail / sizeof *detail), ptr );
delete [] ptr;
}
In the limited version of setting all elements to 0, you can use an extra pair of parenthesis in the new call:
int * ptr = new int[2](); // will value initialize all elements
But you seem to be looking for a different thing.
Given that you're real class is more complex than an int, and constructed from differing values, it's complicated.
A vector can be constructed with iterators if you have an existing array/vector with the correct values to default from, or you have to use placement new.
//vector
int main()
{
int int_static[2] = {1,2};
std::vector<int> int_dynamic(int_static, int_static+2);
//this is what everyone else is saying. For good reason.
}
//placement new
int function_that_returns_constructed_from_values() {
return rand();
}
int main()
{
int count = 2;
char *char_dynamic = new char[count * sizeof(int)];
int *int_dynamic = char_dynamic;
for(int i=0; i<count; ++i)
new(int_dynamic+i)int(function_that_returns_constructed_from_values());
//stuff
for(int i=0; i<count; ++i)
(int_dynamic+i)->~int(); //obviously not really int
delete []char_dynamic;
}
Obviously, the vector is the preferred way to do this.
The initializer data must be somewhere anyway. Simply name it.
E.g.,
#include <stddef.h>
#include <algorithm> // std::copy
#include <vector>
typedef ptrdiff_t Size;
template< class Type, Size n >
Size countOf( Type (&)[n] ) { return n; }
int main()
{
using namespace std;
static int const initData[] = {1,2};
static Size const n = countOf( initData );
// Initialization of a dynamically allocated array:
int* pArray = new int[n];
copy( initData, initData + n, pArray );
// Initialization of a vector:
vector<int> v( initData, initData + n );
}
EDIT: fixed a thinko in above code. I hastened to add example on request. So what I put did erroneously use return value from std::copy.
Cheers & hth.,
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)