When I changed the code to include an argument for size, the code worked correctly. But I cannot understand the error it is showing when I'm calculating length of array using sizeof() ?
#include <iostream>
using namespace std;
void display(int arr[]){
int n=sizeof(arr)/sizeof(arr[0]);
int* ptr=arr;
for (int i=0;i<n;i++) {
cout << i << " : " << *ptr << endl;
ptr++;
}
}
int main(){
int arr[9] = {2,5,3,8,5,4,6,1,7};
display(arr);
return 0;
}
void display(int arr[], int n)
This is a lie, but that's not your fault.
During compilation it is automatically re-written to:
void display(int* arr, int n)
That is, although arrays are not pointers, silly old C features can make us think that they are! It is impossible to pass an array by value so you're really just passing a pointer to the first element.
As such, you're doing sizeof(int*).
Instead, I suggest using sizeof(*arr) * n, or pass a reference to the array:
#include <iostream>
template <std::size_t N>
void display(const int (&arr)[N])
{
std::size_t i = 0;
for (std::size_t i = 0; i < N; i++)
std::cout << i << " : " << arr[i] << '\n';
}
int main(){
int arr[9] = {2,5,3,8,5,4,6,1,7};
display(arr);
}
… or rewrite your function to take an iterator pair instead:
#include <iostream>
#include <iterator>
template <typename It>
void display(const It begin, const It end)
{
std::size_t i = 0;
for (It it = begin; it != end; ++it)
std::cout << i++ << " : " << *it << '\n';
}
int main(){
int arr[9] = {2,5,3,8,5,4,6,1,7};
display(std::begin(arr), std::end(arr));
}
