this my program to return array from function in c++
#include <iostream>
using namespace std;
int *pTest(){
static int a[] = {2,3,4,6,9};
return a;
}
int main(){
int *x;
x = pTest();
while(*x != NULL){
cout << *x++ << " ";
}
}
according to me output should be 2 3 4 6 9
but on my machine output is 2 3 4 6 9 1,
why there is an extra 1 in the output.
i am using codeblocks ,gcc 4.8.1
Arrays aren't zero-terminated, so the loop while (*x != NULL) will keep reading beyond the array until it finds a zero-valued word, or crashes, or causes some other undefined behaviour.
You'll either need to add a terminator to the array (if you can choose a value, perhaps zero, that won't be a valid array element), or return the length in some other way.
You can use a count and take the size of the array.
Like this:
int k = 0;
while(k <= sizeof(x)){
cout << " "<< *x++;
k++;
}
With std::vector your function will be:
std::vector<int> ptest() {
static const int a[] = {2,3,4,6,9};
std::vector<int> vec (a, a + sizeof(a) / sizeof(a[0]) );
return vec;
}
Related
Write a function, reverseArray, that when passed an int array of length greater than 0 will return a dynamically allocated array of the same length but with the elements in the reverse order. For example, if passed the array, {1,2,3,4,5,6,7,8,9,0} the function would return the array {0,9,8,7,6,5,4,3,2,1}.
Below is my code, but there is a bug in it.
This is my output.
1
2
3
4
5
6
4113
6
5
4
3
2
1
0x7fffe697ceb0
The 4113 and address are provided by the compiler.
#include <iostream>
using namespace std;
int * readNumbers() {
int * a = new int[6];
for (int i = 0; i < 6; i++) {
int x;
cin >> x;
a[i] = x;
}
// a++;
return a;
delete[] a;
}
int *reverseArray(int *numbers1,int length) {
for (int i = length; i >=0; i--) {
cout << numbers1[i] << endl;
}
return numbers1;
delete [] numbers1;
}
int main() {
int *arr1 = readNumbers();
cout << reverseArray(arr1,6) << endl;
return 0;
}
I think there may have been an issue with your wording. Assuming you want your function just to print the reverse of a passed array, you're off to a good start.
One issue is what was said in the comments: your for loop is indexing past your array. When you type int * a = new int[6]; you are creating a pointer 'a' which points to a location in memory. Since you chose size 6, the appropriate amount of memory is allocated. If you happen to index outside of that range, you will end up pointing to a random spot in memory, not allocated for your array. Hence why you are getting a weird number '4113'.
A fix for this could be:
int i = length changed to int i = length-1
Another issue is that your function returns an integer pointer, and you are trying to cout this pointer. As another commenter said, you have to think about what this does. If you try this code:
#include <iostream>
using namespace std;
int main() {
int arr[] = {1, 2, 3};
cout << arr << endl;
return 0;
}
your output would be something like 0xff09ba. This represents the location of the start of the array in memory. If you change arr to (arr + 1) you will get the location of the second index of the array.
So when you type cout << reverseArray(arr1,6) << endl; you are really just printing out the location of numbers1 in memory. This is why you are getting '0x7fffe697ceb0' in your output. To fix this, simply make your function
void reverseArray(int *numbers1,int length) {
for (int i = length; i >=0; i--) {
cout << numbers1[i] << endl;
}
}
and change your main to:
int main() {
int *arr1 = readNumbers();
reverseArray(arr1,6);
return 0;
}
Now, if you actually want to return this array, you would need to create a new array which holds the reverse numbers and then return that. An example of a function that does that is:
int* reverseArray(int *numbers1,int length) {
int j = 0;
int *numbers2 = new int[length];
for (int i = length-1; i >=0; i--) {
numbers2[j] = numbers1[i];
j++;
}
return numbers2;
}
There are probably better ways to do this, but this is just one solution. Regardless, you should always be careful when allocating memory yourself.
I have a variable k of type int to set the length of a dynamically allocated int array:
int *Numbers = new int[k];
But because of this I cannot iterate over the array, I get an error:
"no matching begin function was found required for this range-based for statement"
I also cannot get the length of the array using size();
Here's the complete code:
#include <iostream>
using namespace std;
int main()
{
int b, k;
cin >> b >> k;
int *Numbers = new int[k];
for (int i : Numbers) {// (There is a error)
}
for (int i = 0; i < size(Numbers); i++) {
}
}
Prefer using a std::vector instead of a std::array. (Like #tadman mentioned.)
Here is your code using std::vector instead:
#include <iostream>
#include <vector>
int main()
{
int b, k;
std::cin >> b >> k;
std::vector<int> Numbers(b,k); // Fills the vector "Numbers" with nth number of elements with each element as a copy of val.
for (int i : Numbers)
std::cout << i << std::endl;
for (int i = 0; i < Numbers.size(); i++)
std::cout << Numbers[i] << std::endl;
return 0;
}
Say I want 10 elements with the number 5.
Output:
10
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
Also consider not using namespace std;.
The simple and recommended solution is to use std::vector, however if you really want a dynamically allocated array and to use iterator like features on it, you can use iterator_range from boost library, which allows you to create an iterator range for it thus making it usable in range based for loops and in functions like std::size.
Live demo
#include <iostream>
#include<boost/range.hpp>
int main()
{
int k = 5;
int *Numbers = new int[k]{1,4,5,7,8};
auto arr = boost::make_iterator_range(Numbers, Numbers + k);
for (int i : arr) { //range based loop
std::cout << i << " ";
}
std::cout << std::endl << "Size: " << arr.size(); //print size
//or std::size(arr);
}
Output:
1 4 5 7 8
Size: 5
Range-based for loops work with arrays, but not work with pointers. The Actual issue is that arrays is actually a pointer and not an array.try to use simple array.
Using pointers is problematic for many reasons. The simple solution to your problem is to use a vector
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int b, k;
cin >> b >> k;
vector<int> Numbers(k);
for (int i : Numbers) {
cout << i << endl;
}
for (int i = 0; i < Numbers.size(); i++) {
cout << Numbers[i] << endl;
}
}
C array does not have default iterator and thus there is no begin() and end() functions that are used to iterate over array when you use statment like this:
for (int i : Numbers)
You can check range-for reference:
range_expression - any expression that represents a suitable sequence (either an array or an object for which begin and end member functions or free functions are defined, see below) or a braced-init-list.
Okay, so since the dynamic array does not have a default iterator, do not use the for-each loop, instead consider using the regular for loop.
Also, mind the the size function will not work for an array (or dynamic array) and you need to remember the size, since it's not possible to get the size from the pointer only. Hence, this code would work:
#include <iostream>
using namespace std;
int main()
{
int b, k;
cin >> b >> k;
int *Numbers = new int[k];
const int SIZE = k;
for (int i = 0; i < SIZE; i++) {
cout << i << ' ';
}
}
You need to dereference *Numbers by using the * if you want to iterate over the array because *Numbers is a pointer to an integer which points to the first element of your array.For Example :
#include <iostream>
using namespace std;
int main()
{
int k = 10;
int *numbers = new int[k];
//filling the array
for(int i = 0 ; i < k ; ++i) {
*(numbers + i) = i ;
}
//output array element
for(int i = 0 ; i < k ; ++i) {
cout << numbers + i << " is the address of "<<*(numbers + i) << endl;
}
return 0;
}
The output is :
0x6f1750 is the address of 0
0x6f1754 is the address of 1
0x6f1758 is the address of 2
0x6f175c is the address of 3
0x6f1760 is the address of 4
0x6f1764 is the address of 5
0x6f1768 is the address of 6
0x6f176c is the address of 7
0x6f1770 is the address of 8
0x6f1774 is the address of 9
Unfortunatly, you can't get the size of your array with *Numbers because it's not an array but a pointer.
i'm new to programming , this code gives me syntax error in line => int *result = apply_all(array1,5,array2,3) this is the error: expected primary-expression before '}' token|
i'm trying to write function called apply_all expects 2 arrays of integers and their sizes and dynamically allocates a new array of integers whose size is the product of 2 array sizes.
the function should loop through the 2nd array and multiple each element accross each element of array 1 and store the product in newly created array. the function is returning a pointer of to the newly allocated array.
also i wrote a function which is print to display the 1st & 2nd & newly array.
#include <iostream>
using namespace std;
//function prototype
int *apply_all(int *array1 ,int size1,int *array2,int size2);
void print(int *array,int size);
int main()
{
int array1[] {1,2,3,4,5};
int array2[] {10,20,30};
cout << "Array 1:";
print(array1,5);
cout << "Array 2:";
print(array2,3);
int *result = apply_all(array1,5,array2,3);
cout << "Result : ";
print(result,15);
delete [] result;
return 0;
}
int *apply_all(int *array1 ,int size1,int *array2,int size2)
{
int *result {nullptr};
result = new int[size1 * size2];
for (int i{0};i<size2;i++)
for(int j{0};j<size1;j++)
*(result[i*5+j]) = *(array1[i])**(array2[j]);
return result;
}
void print(int *array,int size)
{
for(auto num:array)
cout << num << endl;
}
On this line:
*(result[i*5+j]) = *(array1[i])**(array2[j]);
since result[i*5+j] gives you an int, you are trying to dereference an int, which is not possible.
You just need to do:
result[i*5+j] = array1[i] * array2[j];
Also, in print, your range-for loop won't work with a pointer. You need to do:
for(int i = 0; i < size; ++i)
cout << array[i] << endl;
Also, in apply_all, your loop bounds are incorrect. i needs to go till size1, and j needs to go to size2.
Here's a demo.
Since you are new, a simple work around would be creating an array with buffer space to store your results in and passing the pointer for this into apply_all. You could then write to this array which (being declared in main) should be very easy to access and cause few errors and use a c-string like ending to know when your results are over and to stop printing from the array (c-strings end with a value of 0 so that programs don't read unrelated memory). eg:
int buf[99];
apply_all(array_1, size1, array_2, size2, buf, size3);
for (int x = 0; buf[x] != end of buf var; x++;)
{
print(buf[x])
}
and
apply_all()
{
buf[start-end] = whatever you want;
buf[end + 1] = some variable that won't appear in buffer; //max int size?
}
This should be really simple, but I'm used to higher level languages and am missing something. I'm just trying to make sure the input is five numbers long, and then find the highest number. Unfortunately, something goes wrong in that second part.
#include <iostream>
#include <string>
bool isFiveDigits(int num) {
if (std::to_string(num).length() == 5) {
return true;
} else {
return false;
}
}
int highestInArr(int *nums) {
int highest = nums[0];
for (int i = 1; i < sizeof(nums); i++) {
int temp = nums[i];
if (temp > highest) {
highest = temp;
}
}
return highest;
}
int main() {
using namespace std;
int num;
int nums [5];
cout << "Enter a five digit number!\n";
cin >> num;
if (!isFiveDigits(num)) {
cout << "Not five digits, can you even count?";
return 1;
}
string numstr = to_string(num);
for (int i = 0; i < numstr.length(); i++) {
cout << numstr[i] << " ";
nums[i] = (int)numstr[i];
}
cout << "\n" << highestInArr(nums);
}
When this runs, I get:
Enter a five digit number!
12345
1 2 3 4 5
1424080487
Of course, 1,424,080,487 is not in [1, 2, 3, 4, 5].
You cannot pass a pointer into a function and get the size of it without template deduction. At runtime, all the function receives is a pointer. When you call sizeof(nums), you are not getting the size of the original array. You are simply getting the size of the pointer, which is the same as saying sizeof(int_ptr). Instead, you should be using a std::vector when using collections whose sizes are dynamic.
Now, you CAN receive the size by doing something like this:
#include <iostream>
template<typename num_t, size_t N>
num_t max_num(num_t(&arr)[N]) {
num_t m = (num_t)0;
for (size_t i = 0; i < N; ++i)
if (arr[i] > m)
m = arr[i];
return m;
}
int main(){
int foo[] = { 1, 5, 2, 4, 3 };
int m = max_num(foo);
std::cout << m << std::endl;
std::cin.get();
return 0;
}
However, this is not necessarily preferred and assumes that the array was created on the caller's stack. It does not work for dynamically allocated arrays that were created with new[]. If you do this multiple times with different sizes, you will have multiple implementations of the same function (that's what templates do). The same goes for using an std::array<int, N>. If you use N as a size_t template parameter, it will do the same thing.
There are two preferred options:
Send the size of the array into the function so that the caller is responsible for the size.
Use a different container such as std::vector so the callee is responsible for the size.
Example:
#include <vector>
#include <iostream>
#include <algorithm>
int main(){
std::vector<int> vec{ 1, 5, 2, 4, 3 };
int m = *std::max_element(std::cbegin(vec), std::cend(vec));
std::cout << m << std::endl;
std::cin.get();
return 0;
}
As for the is_5_digits, you should use the base-10 logarithm function.
#include <cmath>
// ...
int i = 12345;
size_t length = (i > 0 ? (int)log10(i) : 0) + 1;
std::cout << length << std::endl; // prints 5;
First of all, you can't simply convert a char to int just like (int)numstr[i] assuming that it will return the digit which it contains.
See, if you have a char '0', it means it's ASCII equivalent is stored, which is 48 in case of 0, 49 in case of '1' and so on.
So in order to get that digit (0,1,2,...,9), you've to substract 48 from the ASCII value.
So change this line:
nums[i] = (int)numstr[i];
to:
nums[i] = (int)numstr[i] - 48; // or nums[i] = (int)numstr[i] - '0';
And another thing, in your highestInArr function, you're getting a pointer as parameter, and in the function, you're using sizeof to determine the size of the array. You can't simply do that, the sizeof will return the size of int*, which is not the size of the array, so you've to pass size as the second argument to the function, and use it in the loop.
Like this:
int highestInArr(int *nums, int size) {
// ...
for (int i = 1; i < size; i++) {
// ...
}
// ...
}
I have a function to which I need to pass an array.
But I don't want to pass the entire array (e.g., valid indices from array[0] to array[size-1]), but a subarray (e.g., valid indices starting at array[5] to array[size-1]).
Is there a way to do that in C++?
You can transfer array to function parameter below
void Foo(int* arr, int length);
//call Foo
Foo(&a[0], length); //or
Foo(a, length);
you can also transfer a certain range of array.
Foo(&a[1], length);
Foo(a + 1, length);
Just, simple code.
#include <iostream>
void Print(int* arr, int length)
{
for(int i=0; i < length; i++)
{
std::cout << *(arr + i) << ", ";
}
std::cout << std::endl;
}
int main()
{
int a[5] = {1,2,3,4,5};
//type A
Print(&a[0], sizeof(a)/sizeof(int)); //print all element of a
Print(&a[1], 3); //print 2,3,4
//type B
Print(a, sizeof(a)/sizeof(int)); //print all element of a
Print(a + 1, 3); //print 2,3,4
getchar();
return 0;
}
Quoted comment by n.m.:
You cannot pass an array to a function. When you try to, you actually pass the address of the first element of the array. If you need a subarray that starts at 5, you just pass the address of the fifth elements. You shouldn't be using C-style arrays anyway. Use std::vector and iterators, this is the C++ way.
As indicated, you can add an offset to the array base pointer (and subtract from the passed arraysize accordingly).
Or pass begin and end (one past the end) pointers of the array to the function to achieve an "iterator-style" implementation.
But as you are programming C++, please consider to use std::vector.
Example:
#include <iostream>
void foo(int arr[], int size) {
for (int i = 0; i < size; i++)
std::cout << arr[i] << ' ';
}
void bar(int* begin, int* end) {
while (begin != end)
std::cout << *begin++ << ' ';
}
int main() {
int arr[] = {0,1,2,3,4,5,6,7,8,9};
int size = sizeof(arr)/sizeof(*arr);
// pass entire array
foo(arr, size);
//bar(arr, arr+size);
std::cout << '\n';
// pass array starting at index 5
foo(arr+5, size-5);
//bar(arr+5, arr+size);
std::cout << '\n';
}
The output is:
$ g++ test.cc && ./a.out
0 1 2 3 4 5 6 7 8 9
5 6 7 8 9