Pseudo Code:
int arr[ 5 ] = { 4, 1, 3, 2, 6 }, x;
x = find(3).arr ;
x would then return 2.
Pseudo Code:
int arr[ 5 ] = { 4, 1, 3, 2, 6 }, x;
x = find(3).arr ;
x would then return 2.
The syntax you have there for your function doesn't make sense (why would the return value have a member called arr
?).
To find the index, use std::distance
and std::find
from the <algorithm>
header.
int x = std::distance(arr, std::find(arr, arr + 5, 3));
Or you can make it into a more generic function:
template <typename Iter>
size_t index_of(Iter first, Iter last, typename const std::iterator_traits<Iter>::value_type& x)
{
size_t i = 0;
while (first != last && *first != x)
++first, ++i;
return i;
}
Here, I'm returning the length of the sequence if the value is not found (which is consistent with the way the STL algorithms return the last iterator). Depending on your taste, you may wish to use some other form of failure reporting.
In your case, you would use it like so:
size_t x = index_of(arr, arr + 5, 3);
the fancy answer. Use std::vector and search with std::find
the simple answer
use a for loop
Here is a very simple way to do it by hand. You could also use the <algorithm>
, as Peter suggests.
#include <iostream>
int find(int arr[], int len, int seek)
{
for (int i = 0; i < len; ++i)
{
if (arr[i] == seek) return i;
}
return -1;
}
int main()
{
int arr[ 5 ] = { 4, 1, 3, 2, 6 };
int x = find(arr,5,3);
std::cout << x << std::endl;
}
#include <vector>
#include <algorithm>
int main()
{
int arr[5] = {4, 1, 3, 2, 6};
int x = -1;
std::vector<int> testVector(arr, arr + sizeof(arr) / sizeof(int) );
std::vector<int>::iterator it = std::find(testVector.begin(), testVector.end(), 3);
if (it != testVector.end())
{
x = it - testVector.begin();
}
return 0;
}
Or you can just build a vector in a normal way, without creating it from an array of ints and then use the same solution as shown in my example.