I have created 2 dynamic arrays in the main function. I have passed both of them to the function by reference. Then I copy data from smaller dynamic array to the larger dynamic array. I delete the smaller dynamic array. Assign the address of the larger dynamic array to the smaller dynamic array. Now ideally the arr array should have size of 10. However, when I try to print the 6th element of the array in the main, it crashes. Please have a look at the code below:
#include <iostream>
#include <string>
using namespace std;
void func(string * arr, string * brr);
int main()
{
string* arr = new string[5];
arr[0] = "hello0";
arr[1] = "hello1";
arr[2] = "hello2";
arr[3] = "hello3";
arr[4] = "hello4";
string* brr = new string[10];
func(arr, brr);
for(int i = 0; i < 6; i++)
cout << arr[i] << endl;
return 0;
}
void func(string * arr, string * brr)
{
for(int i = 0; i < 5; i++)
brr[i] = arr[i];
for(i = 0; i < 5; i++)
cout << brr[i] << endl;
delete []arr;
arr = brr;
arr[5] = "hello5";
}
This line has absolutely no effect for the caller:
arr = brr;
So after the call, arr points exactly where it used to point before - to a now invalid memory area (because you deleted it).
If this would be a C question, I would advise you to use a pointer to a pointer (string **arr). However, I feel this is nasty in a C++ program. Maybe you want to use a reference somewhere ?
Set this signature for the function
void func(string * & arr, string * & brr)
#cnicutar correctly diagnosed the problem; you'll either need to pass arr by reference (reference to the POINTER, not the array), of have fund return the new value of arr, which the caller can assign.
Related
I am trying to implement a dynamically created array using new that I want to resize but the resize action is not working correctly.
This is an exercise on dynamic arrays, so I need dynamic arrays using new and not std::vector.
Here is my code :
int main ()
{
// Build an application here
int length = 0;
int size = 0;
int input;
bool endAdding = false;
cout << "Please enter the length of the new array : ";
cin >> length;
int* oPtrDynamicArray = CreateDynamicArray (length, size);
do
{
cout << "Add an element (0 to quit) : " << endl;
cin >> input;
cout << endl << endl;
if (input == 0){
endAdding = true;
}
else
{
InsertElement(oPtrDynamicArray, input, size, length);
}
cout << "The array contains : " << endl;
for (int i = 0; i < size; i++)
{
cout << i << ": [" << oPtrDynamicArray[i] << "]" << endl;
}
} while (!endAdding);
DeleteDynamicArray (oPtrDynamicArray, size);
}
int *CreateDynamicArray (int capacity, int &size)
{
size = 0;
return new int[capacity];
}
void DeleteDynamicArray (int *dynamicArray, int &size)
{
delete[] dynamicArray;
dynamicArray = nullptr;
size = 0;
}
void InsertElement (int *dynamicArray, int element, int &size, int capacity)
{
if (capacity <= size)
{
ResizeDynamicArray (&dynamicArray, size+1);
}
dynamicArray[size] = element;
size++;
}
void ResizeDynamicArray (int **dynamicArray, int newCapacity)
{
int ** newArray = new int*[newCapacity];
for (int i = 0; i < newCapacity; i++)
{
newArray[i] = dynamicArray[i];
}
*dynamicArray = *newArray;
delete[] newArray;
newArray = nullptr;
}
The problem is that the array is passed to my InsertElement() function and then to ResizeDynamicArray() only if capacity <= size, but the array passed to the first function, is passed with good values, but with abnormal pointers in the array.
Example :
For an array of 3, I have :
array[0] = 1 --> adress 0x0004e300 containing value 1
array[1] = 2 --> adress 0x00000003 containing ???
array[2] = 3 --> adress 0x008ffb24 containing value 2
I really don't understand, it would be really great if someone could explain my error :/.
The problem is here
void InsertElement (int *dynamicArray, int element, int &size, int capacity)
{
if (capacity <= size)
{
ResizeDynamicArray (&dynamicArray, size+1);
}
dynamicArray[size] = element;
size++;
}
when you call ResizeDynamicArray you are changing the dynamicArray pointer declared as a parameter to InsertElement. You are not changing the oPtrDynamicArray pointer in main.
If you want to make this work you need to change InsertElement to take a double pointer (just like ResizeDynamicArray)
void InsertElement (int **dynamicArray, int element, int &size, int capacity)
{
if (capacity <= size)
{
ResizeDynamicArray (dynamicArray, size+1);
}
(*dynamicArray)[size] = element;
size++;
}
Or you could do the easy thing and just use std::vector<int>.
EDIT now that I look at it your ResizeDynamicArray function I see that function is completely wrong as well. It's clear that you have some learning to do with pointers
Here's how ResizeDynamicArray should be
void ResizeDynamicArray (int **dynamicArray, int newCapacity)
{
int * newArray = new int[newCapacity];
for (int i = 0; i < newCapacity; i++)
{
newArray[i] = (*dynamicArray)[i];
}
delete[] *dynamicArray;
*dynamicArray = newArray;
}
You're not the first newbie to fail to understand pointers. Have a good look at the code above and compare it with your code. The main difference is that my code using a pointer to change what is being pointed to. Your code tried to change the pointer itself, which is incorrect. It's confusing because what is being pointed to is another pointer (the dynamic array).
There are several issues in your code:
First, in ResizeDynamicArray, you allocate an array of pointers to ints, not an array of ints. int ** newArray = new int*[newCapacity] should be int *newArray = new int[newCapacity].
Second, once you have fixed that, you need to write *dynamicArray = newArray;;
but you should free the old array before you assign the pointer to the new memory block.
void ResizeDynamicArray (int **dynamicArray, int newCapacity)
{
int *newArray = new int[newCapacity];
for (int i = 0; i < newCapacity; i++)
{
newArray[i] = (*dynamicArray)[i];
}
delete[] *dynamicArray;
*dynamicArray = newArray;
}
Third, you since InsertElement may call ResizeDynamicArray (which will give you back a new memory block), you need to alter the originally passed pointer. So you need to pass a pointer to a pointer int the function, just as you did with ResizeDynamicArray:
void InsertElement (int **dynamicArray, int element, int &size, int capacity)
adapt the body accordingly then.
I know the question is already answered but not the why.
You have to keep in mind that pointers are passed to functions by value.
The pointer value, the address it points to, is lost when the function ends.
But you are still able to change the value stored at the address it points to by dereferencing the pointer.
To pass be able to change the pointer value, the address it points to, inside a function, you have to pass a pointer to a pointer. In this case passing a double pointer is synonym for passing a pointer to a function by reference.
void Foo(int **ptr)
{
// Don't use ptr, changes are lost after function end as ptr is a local copy
// Use *ptr to change the value of the pointer you passed to the function.
// Use **ptr to change the value at the address the pointer you passed to the funcion points to
}
In this case you can change the pointer value, the address it points to, by dereferencing the double pointer once. Which applies to above answers.
I have this function:
void reverse(int* nums, unsigned int size)
This function is supposed to reverse the values in the array it is getting.
Now for reversing I thought to create another array with the size of the array passed in. Assigning this new one from the end of the original array to the start.
But I am a kind of new in C++, So I don't know how to create dynamic array in the size of the parameter of the function.
It's actually not necessary to allocate a new array here. See if you can find a way to solve this problem just by rearranging the existing elements in-place.
Given that this seems like it's an exercise with pointers, you can allocate space by using the new[] operator:
int* auxiliaryArray = new int[size];
You'd then free it by writing
delete[] auxiliaryArray;
However, this isn't the preferred way of doing this in C++. The better route is to use std::vector, which does all its own memory management. That would look like this:
std::vector<int> auxSpace(size);
You can then access elements using the square brackets as you could in a real array. To do this, you'll need to #include <vector> at the top of your program.
In C++, the recommended way to create an array of variable size would be to use an std::vector
#include <vector>
void reverse(int* nums, unsigned int size)
{
std::vector<int> V(size);
...
}
But that approach isn't the best here for performance because it requires additional memory to be allocated of the size of the array, which could be big. It would be better to start from the outside of the array and swap members one by one that are at mirroring positions (so if the size is 5, swap 0 and 4, then swap 1 and 3 and leave 2 alone). This only requires temporary storage of a single int.
You can do it without the need to create another array:
void reverse(int* array, const int size){
for(int i = 0; i < size / 2; i++){
int tmp = array[i];
array[i] = array[size - 1 - i];
array[size - 1 - i] = tmp;
}
}
int main(){
int array[] = {1, 3, 5, 7, 9, 11};
const int size = sizeof(array) / sizeof(array[0]);
reverse(array, size);
for(int i(0); i < size; i++)
std::cout << array[i] << ", ";
}
As you can see above in the loop you only need to swap the first element (element 0) with the n-1 element and the second one with n-1-1 and son on...
Remember arrays are indexed from 0 through n-1.
If you want to allocate new array which is not practical:
int* reverse2(int* array, const int size){
int* tmp = new int[size];
for(int i(size - 1), j(0); j < size; j++, i--)
tmp[j] = array[i];
return tmp;
}
int main(){
int array[] = {1, 3, 5, 7, 9, 11};
for(int i(0); i < size; i++)
std::cout << array[i] << ", ";
std::cout << std::endl;
int* newArray = reverse2(array, size);
for(int i(0) ; i < size; i++)
std::cout << newArray[i] << ", ";
std::cout << std::endl;
delete[] newArray;
return 0;
}
If you want to use a new array you can, but I think is to kill flies with a cannon.
Looks like you are using plain C code and not C++. I say that because of the signature of the function. The signature of the function in a common C++ code could be something like this other:
void reverse(std::vector& items);
You can reverse the current array without a new array, using the current one. You are passing the pointer to the first item of the array, and the content is not constant so that you can modify it. A better signature for the function could be:
void reverse(int* const nums, const unsigned int size);
Looks like a pointer problem. Think about the boundaries to iterate the positions of the array. Would you need to iterate the whole array? Maybe only half array? ;)
As bonus track, what about to exchange the values without an auxiliar variable? (this is true into this case that we are using the fundamental type int... remember the binary arithmetic).
array[pos_head] ^= array[pos_tail];
array[pos_tail] ^= array[pos_head];
array[pos_head] ^= array[pos_tail];
I'm eradicating std::string in favor of C-strings, which I'm new to. How do I get the following to compile? g++ complains: cannot convert char(*)[16] to char**
#include <iostream>
void print(char** s, int n)
{
for (int i = 0; i < n; ++i)
{
std::cout << s[i] << '\n';
}
}
int main()
{
constexpr int n = 3;
char s[n][16]{ "Hello", "Bye", "Sky"};
print(s, n);
}
You created a multidimensional array, not an array of pointers. Usually an array can be said to be equivalent to a pointer, however in this case c++ needs to know the size of the second dimension of your array. The function would be as follows
void print(char s[][16], int n)`{
for (int i = 0; i < n; ++i)
{
std::cout << s[i] << std::endl;
}
}
Understandably you may want to pass the function using pointers as to not make an entire copy of the 2-d array. I saw you mentioned you were okay with variable length strings. That functionality is supported in the string library. You are dealing with c-strings which are not strings at all but static arrays of type character. Defining these c-strings using dynamic memory happens to give you the desired behavior as you create in the simplest terms an array of pointers.
void print(char** s, int n)
{
for (int i = 0; i < n; ++i)
{
std::cout << s[i] << std::endl;
}
}
int main()
{
int n = 3, i;
char** s = new char*[n];
for (i = 0; i < 3; i++) {
s[i] = new char[16];
}
s[0] = "Hello";
s[1] = "Bye";
s[2] = "Sky";
print(s, n);
for (i = 0; i < 3; i++) {
delete [] s[i];
}
delete [] s;
s = NULL;
return 0;
}
Since you are using dynamic memory now you need to free your memory which is what the last loop serves to do. As you can see using all this dynamic memory is quite taxing and it would be easier to use the string library that has been optimized to do a much better job then you can. If you're still not convinced you should at least make your own string class to handle the dynamic memory that contains a char * as its private member. In either case you avoid this mess and just make an array of zed class objects and not deal at all with multidimensional nonsense. No one likes seg faults and memory leaks.
Given any type T, T arr[N]; declares a variable arr of type T[N], which is an array and not a pointer. When you use arr in almost all contexts, array to pointer conversions happen, giving the incorrect illusion that arr is a pointer of type T*.
char s[n][16] = { "Hello", "Bye", "Sky" };
declares s as an array of n elements of type char[16]. Now, when array to pointer conversion happens, s decays into a pointer of type char (*)[16]. Hence, your function needs to have the signature
void print(char (*s)[16], int n);
Which is equivalent to
void print(char s[][16], int n);
the [] is interpreted as a pointer by the compiler.
To make these complex types more readable, a type alias may be used.
using T = char[16];
void print(T s[], int n);
Addressing some concerns
As pointed out in the comments, std::string should almost always be preferred over a char array. If you have performance concerns, benchmark before doing this. I really doubt much performance gains can be observed in most cases.
Declaring an array with length n which is an int is not standard C++. It is an extension provided by your compiler, it is not portable and in most cases not necessary.
int n = 3;
char vla[n]; // this is a variable length array
char arr[3]; // this is just an array
char* darr = new char[3]; // this is a pointer pointing to dynamically allocated memory
std::string str; // but instead, this is just better
The compiler cannot extract from char ** the infomation about char[16]. You need to define a type char[16] and pass the pointer to this type to your print function.
#include <iostream>
typedef char str_t[16];
void print(str_t* s, int n)
{
for (int i = 0; i < n; ++i)
{
std::cout << s[i] << std::endl;
}
}
int main()
{
int n = 3;
char s[n][16]{ "Hello", "Bye", "Sky"};
print(s, 3);
}
I need to implement a function that modifies an array. The new array may be a different size. cout prints 1. I understand what's wrong with this code but I just cannot figure out what the syntax is.
//tried this..
int reduce(int *array[])
{
*array = new int[1];
(*array)[0] = 6;
return 0;
}
//also tried this..
int reduce(int array[])
{
array = new int [1];
array[0] = 6;
return 0;
}
int main()
{
int a[1] = {1};
int *p = a;
reduce(&p);
cout << a[0];
return 0;
}
Don't understand your question correctly, but this is what you may do:
void reduce(int *a, int size)
{
for (int i =0; i < size; ++i) {
*(a+i) = 6; // or whatever value to want to
}
}
Call it this way:
int main(){
int a[5] = {1, 1, 1, 1, 1};
int *p = a;
reduce(p, 5);
for (int i =0; i < 5; ++i) { cout << a[i]<<endl; }
return 0;
}
EDIT
What you are trying to do can be vaguely done this way:
int * reduce (int **b, int size) {
*b = new int[size];
for (int i =0; i < size; ++i) {
*(*b + i) = 6;
}
return *b;
}
int main(){
int a[5] = {1, 1, 1, 1, 1};
int *p = a;
p = reduce(&p, 5);
cout << p[0];
cout << p[1];
cout << p[2];
cout << p[3];
cout << p[4];
delete [] p;
return 0;
}
But it still wont change where a is pointing to.
What you are trying to do is not possible with statically defined arrays.
When you use an array like
int a[1] = {1};
you cannot change the size of the array at run time, you cannot make it point to dynamically allocated memory. You may only access and modify the elements the array. That's it.
The function reduce changes where p points to but it does not change the elements of a.
If you want to modify the contents of a, you can simply use a as an argument, and set the values.
MODIFIED:
You want to modify array a, try this :
int reduce(int **array)
{
*array = new int[1];
(*array)[0] = 6;
return 0;
}
int main()
{
int *a = new int[1];
reduce(&a);
cout << a[0];
return 0;
}
First of all, the formal parameter int* array[] actually is the same as int** array (you can think of it as a two-dimensional array). This is probably not what you want.
The answer of #everettjf will only work if you do not change the size of the array. A possible solution (that completely replaces the array) would be
#include <iostream>
void print_array(int[],int);
int* reduce(int array[]) {
// get rid of the old array
delete[] array;
// create a new one
array = new int[7]{8,4,6,19,3,56,23};
// need to return the new address, so that
// the main function is informed on the new
// address
return array;
}
int main() {
// initialize array
int *a = new int[1]{4};
print_array(a,1);
// "change" array by completely replacing it
a=reduce(a);
print_array(a,7);
return 0;
}
// simply prints out the array; no error checking!
void print_array(int array[], int length) {
std::cout << "[";
for (int i = 0; i < length ; ++i) {
std::cout << array[i] << " ";
}
std::cout << "]" << std::endl;
}
In the reduce function, the initial array is completely deleted. Afterwards, you can create a new one (I chose to just use 7 random numbers). It is important to return that pointer back to the caller (the main method). Otherwise the a pointer in the main method would point to invalid
If you are not forced (by some kind of excercise, for example) to use arrays, you should look into http://en.cppreference.com/w/cpp/container/vector
The premise of your question is invalid. It is not possible to resize an array of automatic storage duration (aka a in main()) after its definition by ANY means in standard C++.
Dynamic memory allocations in either of your reduce() functions will not cause a in main() to be resized.
reduce(&p) will calls the first version of reduce() , which will then change p (so it points at the dynamically allocated memory) but not affect a.
If main() calls reduce(a) or reduce(p) (the two are equivalent, given the initialisation int *p = a) will change neither a nor p, but instead cause a memory leak.
The underlying problem, I suspect, is that you believe - incorrectly - that pointers and arrays are the same thing. They are actually different things, but can be used in the same way in various contexts. And your code is one of the contexts in which they cannot be used interchangeably.
If you want a resizeable array, use a static container (like std::vector<int>) and - if you want a function to resize it, pass it by reference. It manages its own memory dynamically, so is able to dynamically resize itself.
I am trying to code a class that represents a set of integers. It's a homework assignment but for the life of me I cannot figure out this issue.
In the class "IntSet", I have two private variables; one is a pointer to an array the other is the size of the array. I can create objects of this class and they work as intended. But I have this function named "join" that returns an object of the IntSet class. It essentially concatenates the arrays together then uses that array to create the returning object.
Here is my code:
#include <iostream>
using namespace std;
class IntSet {
int * arrPtr;
int arrSize;
public:
//Default Constructor
IntSet() {
int arr[0];
arrPtr = arr;
arrSize = 0;
}
//Overloaded Constructor
IntSet(int arr[], int size) {
arrPtr = arr;
arrSize = size;
}
//Copy Constructor
IntSet(const IntSet &i) {
arrPtr = i.arrPtr;
arrSize = i.arrSize;
}
/*
* Returns a pointer to the first
* element in the array
*/
int* getArr() {
return arrPtr;
}
int getSize() {
return arrSize;
}
IntSet join(IntSet &setAdd) {
//Make a new array
int temp[arrSize + setAdd.getSize()];
//Add the the values from the current instance's array pointer
//to the beginning of the temp array
for (int i = 0; i < arrSize; i++) {
temp[i] = *(arrPtr + i);
}
//Add the values from the passed in object's array pointer
//to the temp array but after the previously added values
for (int i = 0; i < setAdd.getSize(); i++) {
temp[i + arrSize] = *(setAdd.getArr() + i);
}
//Create a new instance that takes the temp array pointer and the
//size of the temp array
IntSet i(temp, arrSize + setAdd.getSize());
//Showing that the instance before it passes works as expected
cout << "In join function:" << endl;
for (int j = 0; j < i.getSize(); j++) {
cout << *(i.getArr() + j) << endl;
}
//Return the object
return i;
}
};
int main() {
//Make two arrays
int arr1[2] = {2 ,4};
int arr2[3] = {5, 2, 7};
//Make two objects normally
IntSet i(arr1, 2);
IntSet j(arr2, 3);
//This object has an "array" that has arr1 and arr2 concatenated, essentially
//I use the copy constructor here but the issue still occurs if I instead use
//Inset k = i.join(j);
IntSet k(i.join(j));
//Shows the error. It is not the same values as it was before it was returned
cout << "In main function:" << endl;
for (int l = 0; l < k.getSize(); l++) {
cout << *(k.getArr() + l) << endl;
}
return 0;
}
The program compiles and the output as of now is:
In join function:
2
4
5
2
7
In main function:
10
0
-2020743083
32737
-2017308032
I don't know why but the 10 and 0 are always the same every time I recompile and run. Also, if I print out the address of the pointer rather than the value(in both the join function and the main function), I get the same memory address.
Sorry if I misuse terms, I come from a java background, so pointers and such are a little new to me. If any clarification is needed, please ask.
Thanks in advance.
int temp[arrSize + setAdd.getSize()];
This is a local array, its lifetime ends once the function returned.
IntSet i(temp, arrSize + setAdd.getSize());
Here you are constructing an IntSet with this array. In fact the constructor simply changes a member pointer to the value of temp:
IntSet(int arr[], int size) {
arrPtr = arr;
arrSize = size;
}
As a result, since the lifetime of the object that temp and consequently also i.arrPtr is pointing to ends after leaving join, you will have a wild pointer. Dereferencing this pointer later in main invokes undefined behavior.
You need to allocate the array dynamically with new[] and delete it later with delete[]. The same goes for your constructors. Also note that if you use new[] in join and delete[] in the destructor, then you also have to make sure that the copy constructor actually copies the array (create new array with new[] and copy contents). If you simply assign the pointer then both the source and destination object will point to the same array and they will also both try to delete it at deconstruction, again invoking undefined behaviour.
But since this C++, you might as well use a std::vector which does all of this for you. (or std::set if you actually want a integer set)
The quickest fix with your code is to change
int temp[arrSize + setAdd.getSize()];
into this
int * temp = new int[arrSize + setAdd.getSize()];
The thing is that you allocated temp on the stack, so when join() returns that memory is releases. By allocating memory on the heap (as per the fix) the memory is not released when join() returns.
There are other things wrong with your code -- depending on the point of the assignment. I think most of these will be fixed when you consider the implications of having memory on the heap.