Related
I am basically looking for some sort of "dynamic" way of passing the size/length of an array to a function.
I have tried:
void printArray(int arrayName[])
{
for(int i = 0 ; i < sizeof(arrayName); ++i)
{
cout << arrayName[i] << ' ';
}
}
But I realized it only considers its bytesize and not how many elements are on the array.
And also:
void printArray(int *arrayName)
{
while (*arrayName)
{
cout << *arrayName << ' ';
*arrayName++;
}
}
This has at least printed me everything but more than what I expected, so it doesn't actually work how I want it to.
I reckon it is because I don't exactly tell it how big I need it to be so it plays it "safe" and throws me some big size and eventually starts printing me very odd integers after my last element in the array.
So I finally got this work around, yet I believe there is something better out there!:
void printArray(int *arrayName)
{
while (*arrayName)
{
if (*arrayName == -858993460)
{
break;
}
cout << *arrayName << ' ';
*arrayName++;
}
cout << '\n';
}
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460, so I made it break the while loop once this value is encountered.
include <iostream>
include <conio.h>
using namespace std;
// functions prototypes
void printArray (int arrayName[], int lengthArray);
// global variables
//main
int main ()
{
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray,3);
printArray (secondArray,5);
// end of program
_getch();
return 0;
}
// functions definitions
void printArray(int arrayName[], int lengthArray)
{
for (int i=0; i<lengthArray; i++)
{
cout << arrayName[i] << " ";
}
cout << "\n";
}
Thank you very much.
TL;DR answer: use std::vector.
But I realized it [sizeof()] only considers its bytesize and not how many elements are on the array.
That wouldn't be a problem in itself: you could still get the size of the array using sizeof(array) / sizeof(array[0]), but the problem is that when passed to a function, arrays decay into a pointer to their first element, so all you can get is sizeof(T *) (T being the type of an element in the array).
About *arrayName++:
This has at least printed me everything but more than what I expected
I don't even understand what inspired you to calculate the size of the array in this way. All that this code does is incrementing the first object in the array until it's zero.
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460
That's a terrible assumption and it also relies on undefined behavior. You can't really be sure what's in the memory after the first element of your array, you should not even be accessing it.
Basically, in C++, if you want to know the size of a raw array from within a function, then you have to keep track of it manually (e. g. adding an extra size_t size argument), because of the way arrays are passed to functions (remember, they "decay into" a pointer). If you want something more flexible, consider using std::vector<int> (or whatever type of objects you want to store) from the C++ standard library -- it has a size() method, which does exactly what you want.
1st try
When arrays are passed into functions they decay to pointers. Normally, using sizeof on an array would give you its size in bytes which you could then divide by the size in bytes of each element and get the number of elements. But now, since you have a pointer instead of an array, calling sizeof just gives you the size of the pointer (usually 4 or 8 bytes), not the array itself and that's why this fails.
2nd try
The while loop in this example assumes that your array ends with a zero and that's very bad (unless you really did use a zero as a terminator like null-terminated strings for example do). If your array doesn't end with a zero you might be accessing memory that isn't yours and therefore invoking undefined behavior. Another thing that could happen is that your array has a zero element in the middle which would then only print the first few elements.
3rd try
This special value you found lurking at the end of your array can change any time. This value just happened to be there at this point and it might be different another time so hardcoding it like this is very dangerous because again, you could end up accessing memory that isn't yours.
Your final code
This code is correct and passing the length of the array along with the array itself is something commonly done (especially in APIs written in C). This code shouldn't cause any problems as long as you don't pass a length that's actually bigger than the real length of the array and this can happen sometimes so it is also error prone.
Another solution
Another solution would be to use std::vector, a container which along with keeping track of its size, also allows you to add as many elements as you want, i.e. the size doesn't need to be known at runtime. So you could do something like this:
#include <iostream>
#include <vector>
#include <cstddef>
void print_vec(const std::vector<int>& v)
{
std::size_t len = v.size();
for (std::size_t i = 0; i < len; ++i)
{
std::cout << v[i] << std::endl;
}
}
int main()
{
std::vector<int> elements;
elements.push_back(5);
elements.push_back(4);
elements.push_back(3);
elements.push_back(2);
elements.push_back(1);
print_vec(elements);
return 0;
}
Useful links worth checking out
Undefined behavior: Undefined, unspecified and implementation-defined behavior
Array decay: What is array decaying?
std::vector: http://en.cppreference.com/w/cpp/container/vector
As all the other answers say, you should use std::vector or, as you already did, pass the number of elements of the array to the printing function.
Another way to do is is by putting a sentinel element (a value you are sure it won't be inside the array) at the end of the array. In the printing function you then cycle through the elements and when you find the sentinel you stop.
A possible solution: you can use a template to deduce the array length:
template <typename T, int N>
int array_length(T (&array)[N]) {
return N;
}
Note that you have to do this before the array decays to a pointer, but you can use the technique directly or in a wrapper.
For example, if you don't mind rolling your own array wrapper:
template <typename T>
struct array {
T *a_;
int n_;
template <int N> array(T (&a)[N]) : a_(a), n_(N) {}
};
You can do this:
void printArray(array<int> a)
{
for (int i = 0 ; i < a.n_; ++i)
cout << a.a_[i] << ' ';
}
and call it like
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray);
printArray (secondArray);
The key is that the templated constructor isn't explicit so your array can be converted to an instance, capturing the size, before decaying to a pointer.
NB. The wrapper shown isn't suitable for owning dynamically-sized arrays, only for handling statically-sized arrays conveniently. It's also missing various operators and a default constructor, for brevity. In general, prefer std::vector or std::array instead for general use.
... OP's own attempts are completely addressed elsewhere ...
Using the -858993460 value is highly unreliable and, in fact, incorrect.
You can pass a length of array in two ways: pass an additional parameter (say size_t length) to your function, or put a special value to the end of array. The first way is preferred, but the second is used, for example, for passing strings by char*.
In C/C++ it's not possible to know the size of an array at runtime. You might consider using an std::vector class if you need that, and it has other advantages as well.
When you pass the length of the array to printArray, you can use sizeof(array) / sizeof(array[0]), which is to say the size in bytes of the whole array divided by the size in bytes of a single element gives you the size in elements of the array itself.
More to the point, in C++ you may find it to your advantage to learn about std::vector and std::array and prefer these over raw arrays—unless of course you’re doing a homework assignment that requires you to learn about raw arrays. The size() member function will give you the number of elements in a vector.
In C/C++, native arrays degrade to pointers as soon as they are passed to functions. As such, the "length" parameter has to be passed as a parameter for the function.
C++ offers the std::vector collection class. Make sure when you pass it to a function, you pass it by reference or by pointer (to avoid making a copy of the array as it's passed).
#include <vector>
#include <string>
void printArray(std::vector<std::string> &arrayName)
{
size_t length = arrayName.size();
for(size_t i = 0 ; i < length; ++i)
{
cout << arrayName[i] << ' ';
}
}
int main()
{
std::vector<std::string> arrayOfNames;
arrayOfNames.push_back(std::string("Stack"));
arrayOfNames.push_back(std::string("Overflow"));
printArray(arrayOfNames);
...
}
I am trying to create an array that generates random values, then assign a pointer to that array in order to use it in other functions.
Question 1: Is this the right approach?
Question 2: When I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
int size = 100;
int theray[size];
for(int i=0; i<size; i++)
{
theray[i] = (rand()%100);
}
//Output array
cout<<"The array: ";
for(int j=0; j<size; j++)
{
cout<<theray[j]<<" ";
}
cout<<endl;
int (*parray)[100] = &theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<*parray[k]<<" ";
}
Question 1: is this the right approach?
No. The right approach is to use std::vector<int> if size is not known at compile time1, and std::array<int, size> if it is2. There is no need for pointers here.
void foo(const std::vector<int>& v)
{
// do stuff with v
}
...
std::vector<int> v(size); // vector with size elements
// do something with v
// pass v to a function
foo(v);
Question 2: when I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
If you use C++ idioms you won't even encounter this problem, so I consider the question moot. However, in your case you have a problem of operator precedence: be explicit about applying de-reference * before access []:
cout<< (*parray)[k] << " ";
1 As shown in the example, you can use an std::vector as a fixed size array, where the size need not be known at runtime. Just bear in mind that it is possible to change it's size after construction.
2In your example, size is not a compile time constant so you cannot use std::array. However, if you had declared it as const int size = 100; then it would be considered a compile time constant.
Your code is a bit off in three ways. First, there is no need to use &theray. Array names already reference a memory address. You can simply assign the pointer to theray. Second, you're declaring an array of 100 pointers. Based on your description, it sounds like you just want one pointer that points to the array. Your declaration should just be int *parray instead of int *parray [100]. Finally, once you have a pointer to the array, you can access elements of the array the same way you would with the original array, only with the name of the pointer, instead of the name of the array. Try changing your last block of code (starting with the pointer declaration to this:
int *parray;
parray = theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<parray[k]<<" ";
}
Question 1
Is this the right approach?
Usually not. It depends on what you are trying to achieve.
For high level semantics you'd in most cases use std::vector<int> or, if the size is fixed and you are using C++11, std::array<int, size>. If you actually have to go down to the pointer level, you'd usually write it like this:
int *parray = theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<parray[k]<<" ";
}
This works because arrays will degrade to pointers, and the […] subscripts work on these pointers just like they work on the original arrays.
Question 2
When I run the code below, my pointer function generates values inconsistent with what the actual array's value is, what am I doing wrong?
*parray[k] gets interpreted as *(parray[k]) while you intend to use it as (*parray)[k].
Question 1: is this the right approach?
No. Use std::vector<> for arrays whose size can change dynamically (at run-time). Prefer avoiding pointers and manual memory management.
Question 2: when I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
First of all, the fact of creating pointers so you can pass the array to a function. This is not necessary. Here is how I would use classes from the C++ Standard Library to write that program (in C++11):
#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
// Sample function that prints the vectors's content
void foo(std::vector<int> const& v)
{
copy(begin(v), end(v), std::ostream_iterator<int>(std::cout, " "));
}
int main()
{
// Populate the vector...
size_t sz = 10;
std::vector<int> v(sz);
generate(begin(v), end(v), [] () { return rand() % 100; });
// Pass it to a function...
foo(v);
}
here is my code and im not allowed to use a loop in the subarray function im pretty confused maybe someone can point me in the right direction i feel like im almost there..
int *duplicateArray(int *arr, int size)
{
int *newArray;
if (size<=0)
return NULL;
newArray = new int[size];
for (int index=0;index<size;index++)
newArray[index]=arr[index];
return newArray;
}
int* subArray(int *sub, int start, int length)
{
int aa[10]={1,2,3,4,5,6,7,8,9,10};
int *dup;
dup = aa;
duplicateArray(dup,10);
return dup;
}
int main()
{ cout << "Testing subArray: " << endl
<< "Expected result: 5, 6, 7, 8 " << endl;
int *subArr;
int start = 5;
subArr = subArray(subArr, 5,4);
for (int index = start; index<10; index++)
cout << subArr[index];
delete [] subArr;
subArr = 0;
So, since this is homework, I'm going to avoid posting a solution directly. You say that;
im not allowed to use a loop in the subarray function
Yet, currently, subArray calls duplicateArray, which uses a loop. This seems to be in conflict with the spirit of the requirement.
Since you haven't said otherwise, I'm assuming that subArray should duplicate the contents of its argument between start and the end. So, what do we know?
We know that the size of the returned array should be length - start elements. We also know (well, perhaps) that a function named memcpy exists which allows you to copy a block of bytes from one place to another (assuming they do not overlap).
(note that I am suggesting memcpy here because we are dealing with POD types (Plain Old Data) and because I doubt your class has delved into the STL. In the future you will be better served by something like std::copy(), but for now this is ok)
So, in order, we need to:
Declare a new array to return with length - start elements. You must dynamically allocate this array! Currently you are returning a pointer to a locally declared array. That pointer becomes invalid as soon as the function returns.
Copy length - start elements (elements, not bytes! Make sure to take into account the number of elements as well as the size of an individual element) from sub + start into this new array.
Return the new array (pointer really).
If I have somehow violated the requirements or intent of your assignment then you need to elaborate on your problem for me. Currently there is not much to go on.
Following is my code which doesnot properly bubbles up the larger value .Can any one help out .The problem is somewhat at count++ .
#include<iostream>
using namespace std;
class heap{
public:
int count;
heap(int c)
{
count=c;
}
int Arr[10];
void insert(int num);
void deletemax();
void print();
};
void heap::insert(int num){
if(count==10){
cout<<"Heap full\n";
exit(1);
}
else{
Arr[count]=num;
count++; //The real problem arises here that the compiler adds 1 to count and when the code moves ahead it sets position var to count++ value and tries to compare a value at Arr[POS] with its parent whereas there is no value at this place set uptill.
}
int POS=count;
while(Arr[POS]>Arr[(POS-1)/2]){
int temp;
temp=Arr[POS];
Arr[(POS-1)/2]=temp;
POS=(POS-1)/2;
}
}
void heap::print(){
for(int i=0; i<10; i++){
cout<<Arr[i]<<endl;
}
}
int main(){
heap h(0);
int a;
int b=0;
while(b<10){
cout<<"Insert node in heap\n";
cin>>a;
h.insert(a);
b++;
}
h.print();
return 0;
}
I would agree, that's where your problem is.
There are many issues with the code you posted, some of which include:
As to your specific issue, I would guess you need to change the line in heap::insert() to "int POS=count-1;" to properly start iterating from the back of the array.
You need to consider the case of adding an element to an empty array and what then happens in your sorting code.
Your constructor allows someone to create a heap that will overflow the fixed sized array, for example heap(1000). In addition, the Arr member is not initialized which means it has undefined data for any value but heap(0). In this case your constructor should not take any parameters and count should just be initialized to 0.
The purpose of the code is confusing. Is it a heap, a sorted array, an attempt to approximate a heap with an array, none of the above? If you are simply trying to implement a fixed sized sorted array then I believe your sorting code in insert() will not work (e.g., consider adding 100 to a heap containing [1,2,3]).
There are other, more basic things, wrong (like not using any the STL containers, public class member, non-const parameter passing, "using std", etc...) but I'm assuming you are merely experimenting/playing here.
Let's say I have a function called MyFunction(int myArray[][]) that does some array manipulations.
If I write the parameter list like that, the compiler will complain that it needs to know the size of the array at compile time. Is there a way to rewrite the parameter list so that I can pass an array with any size to the function?
My array's size is defined by two static const ints in a class, but the compiler won't accept something like MyFunction(int myArray[Board::ROWS][Board::COLS]).
What if I could convert the array to a vector and then pass the vector to MyFunction? Is there a one-line conversion that I can use or do I have to do the conversion manually?
In C++ language, multidimensional array declarations must always include all sizes except possibly the first one. So, what you are trying to do is not possible. You cannot declare a parameter of built-in multidimensional array type without explicitly specifying the sizes.
If you need to pass a run-time sized multidimensional array to a function, you can forget about using built-in multidimensional array type. One possible workaround here is to use a "simulated" multidimensional array (1D array of pointers to other 1D arrays; or a plain 1D array that simulates multidimensional array through index recalculation).
In C++ use std::vector to model arrays unless you have a specific reason for using an array.
Example of a 3x2 vector filled with 0's called "myArray" being initialized:
vector< vector<int> > myArray(3, vector<int>(2,0));
Passing this construct around is trivial, and you don't need to screw around with passing length (because it keeps track):
void myFunction(vector< vector<int> > &myArray) {
for(size_t x = 0;x < myArray.length();++x){
for(size_t y = 0;y < myArray[x].length();++y){
cout << myArray[x][y] << " ";
}
cout << endl;
}
}
Alternatively you can iterate over it with iterators:
void myFunction(vector< vector<int> > &myArray) {
for(vector< vector<int> >::iterator x = myArray.begin();x != myArray.end();++x){
for(vector<int>::iterator y = x->begin();y != x->end();++y){
cout << *y << " ";
}
cout << endl;
}
}
In C++0x you can use the auto keyword to clean up the vector iterator solution:
void myFunction(vector< vector<int> > &myArray) {
for(auto x = myArray.begin();x != myArray.end();++x){
for(auto y = x->begin();y != x->end();++y){
cout << *y << " ";
}
cout << endl;
}
}
And in c++0x for_each becomes viable with lambdas
void myFunction(vector< vector<int> > &myArray) {
for_each(myArray.begin(), myArray.end(), [](const vector<int> &x){
for_each(x->begin(), x->end(), [](int value){
cout << value << " ";
});
cout << endl;
});
}
Or a range based for loop in c++0x:
void myFunction(vector< vector<int> > &myArray) {
for(auto x : myArray){
for(auto y : *x){
cout << *y << " ";
}
cout << endl;
}
}
*I am not near a compiler right now and have not tested these, please feel free to correct my examples.
If you know the size of the array at compile time you can do the following (assuming the size is [x][10]):
MyFunction(int myArray[][10])
If you need to pass in a variable length array (dynamically allocated or possibly just a function which needs to take different sizes of arrays) then you need to deal with pointers.
And as the comments to this answer state:
boost::multiarray may be appropriate since it more efficiently models a multidimensional array. A vector of vectors can have performance implications in critical path code, but in typical cases you will probably not notice an issue.
Pass it as a pointer, and take the dimension(s) as an argument.
void foo(int *array, int width, int height) {
// initialize xPos and yPos
assert(xPos >= 0 && xPos < width);
assert(yPos >= 0 && yPos < height);
int value = array[yPos * width + xPos];
}
This is assuming you have a simple two-dimensional array, like int x[50][50].
There are already a set of answers with the most of the common suggestions: using std::vector, implementing a matrix class, providing the size of the array in the function argument... I am only going to add yet another solution based on native arrays --note that if possible you should use a higher level abstraction.
At any rate:
template <std::size_t rows, std::size_t cols>
void function( int (&array)[rows][cols] )
{
// ...
}
This solution uses a reference to the array (note the & and the set of parenthesis around array) instead of using the pass-by-value syntax. This forces the compiler not to decay the array into a pointer. Then the two sizes (which could have been provided as compile time constants can be defined as template arguments and the compiler will deduct the sizes for you.
NOTE: You mention in the question that the sizes are actually static constants you should be able to use them in the function signature if you provide the value in the class declaration:
struct test {
static const int rows = 25;
static const int cols = 80;
};
void function( int *array[80], int rows ) {
// ...
}
Notice that in the signature I prefer to change the double dimension array for a pointer to an array. The reason is that this is what the compiler interprets either way, and this way it is clear that there is no guarantee that the caller of the function will pass an array of exactly 25 lines (the compiler will not enforce it), and it is thus apparent the need for the second integer argument where the caller passes the number of rows.
You can't pass an arbitrary size like that; the compiler doesn't know how to generate the pointer arithmetic. You could do something like:
MyFunction(int myArray[][N])
or you could do:
MyFunction(int *p, int M, int N)
but you'll have to take the address of the first element when you call it (i.e. MyFunction(&arr[0][0], M, N).
You can get round all of these problems in C++ by using a container class; std::vector would be a good place to start.
The compiler is complaining because it needs to know the size of the all but the first dimension to be able to address an element in the array. For instance, in the following code:
int array[M][N];
// ...
array[i][j] = 0;
To address the element, the compiler generates something like the following:
*(array+(i*N+j)) = 0;
Therefore, you need to re-write your signature like this:
MyFunction(int array[][N])
in which case you will be stuck with a fixed dimension, or go with a more general solution such as a (custom) dynamic 2D array class or a vector<vector<int> >.
Use a vector<vector<int> > (this would be cheating if underlying storage was not guaranteed to be contiguous).
Use a pointer to element-of-array (int*) and a size (M*N) parameter. Here be dragons.
First, lets see why compiler is complaining.
If an array is defined as int arr[ ROWS ][ COLS ]; then any array notation arr[ i ][ j ] can be translated to pointer notation as
*( arr + i * COLS + j )
Observe that the expression requires only COLS, it does not require ROWS. So, the array definition can be written equivalently as
int arr [][ COLS ];
But, missing the second dimension is not acceptable. For little more details, read here.
Now, on your question:
Is there a way to rewrite the
parameter list so that I can pass an
array with any size to the function?
Yes, perhaps you can use a pointer, e.g. MyFunction( int * arr );. But, think about it, how would MyFunction() know where to stop accessing the array? To solve that you would need another parameter for the length of the array, e.g. MyFunction( int * arr, size_t arrSize );
Yes: MyFunction(int **myArray);
Careful, though. You'd better know what you're doing. This will only accept an array of int pointers.
Since you're trying to pass an array of arrays, you'll need a constant expression as one of the dimentions:
MyFunction(int myArray[][COLS]);
You'll need to have COLS at compile time.
I suggest using a vector instead.
Pass a pointer and do the indexing yourself or use a Matrix class instead.
yes - just pass it as pointer(s):
MyFunction(int** someArray)
The downside is that you'll probably need to pas the array's lengths as well
Use MyFunction(int *myArray[])
If you use MyFunction(int **myArray) an pass int someArray[X][Y], the program will crash.
EDIT: Don't use the first line, it's explained in comments.
I don't know about C++, but the C99 standard introduced variable length arrays.
So this would work in a compiler that supports C99:
void func(int rows, int cols, double[rows][cols] matrix) {
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
printf("%f", matrix[r][c]);
}
}
}
Note that the size arguments come before the array. Really, only the number of columns has to be known at compile time, so this would be valid as well:
void func(int rows, int cols, double[][cols] matrix)
For three or more dimensions, all but the first dimension must have known sizes. The answer ArunSaha linked to explains why.
Honestly, I don't know whether C++ supports variable-length arrays, so this may or may not work. In either case, you may also consider encapsulating your array in some sort of matrix class.
EDIT: From your edit, it looks like C++ may not support this feature. A matrix class is probably the way to go. (Or std::vector if you don't mind that the memory may not be allocated contiguously.)
Don't pass an array, which is an implementation detail. Pass the Board
MyFunction(Board theBoard)
{
...
}
in reality my array's size is defined by two static const ints in a class, but the compiler won't accept something like MyFunction(int myArray[Board::ROWS][Board::COLS]).
That's strange, it works perfectly fine for me:
struct Board
{
static const int ROWS = 6;
static const int COLS = 7;
};
void MyFunction(int myArray[Board::ROWS][Board::COLS])
{
}
Maybe ROWS and COLS are private? Can you show us some code?
In C++, using the inbuilt array types is instant fail. You could use a boost::/std:: array of arrays or vector of arrays. Primitive arrays are not up to any sort of real use
In C++0x, you can use std::initializer_list<...> to accomplish this:
MyFunction(std::initializer_list<std::initializer_list<int>> myArray);
and use it (I presume) like this (with the range based for syntax):
for (const std::initializer_list<int> &subArray: myArray)
{
for (int value: subArray)
{
// fun with value!
}
}