template function to find subarray in C++ - c++

Question

#include <iostream>
#include <string>
using namespace std;
template <class T> int IsSubArr(T& a, int a_len, T& b, int b_len)
{
int i,j;
bool found;
int k;
T& s=a,l=b;
int s_len = (a_len < b_len) ? a_len : b_len; // find the small array length
if (s_len == a_len) // check to set pointers to small and long array
{
s = a;
l = b;
}
else
{
s = b;
l = a;
}
for (i = 0; i <= a_len-s_len; i++) //loop on long array
{
found = true;
k=i;
for (j=0; j<s_len; j++) // loop on sub array
{
if (s[j] != l[i])
{
found = false;
break;
}
k++;
}
}
if (found)
return i;
else
return -1;
}
/******* main program to test templates ****/
int main()
{
int array[5] = {9,4,6,2,1};
int alen = 5;
int sub_arr[3] = {6,2,1};
int slen = 3;
int index= 0;
index = IsSubArr(array,alen,sub_arr,slen);
cout << "\n\n Place of sub array in long array: " << index;
cout << endl;
return 0;
}
for this line of code:
index = IsSubArr(array,alen,sub_arr,slen);
i get error:
Error 1 error C2782: 'int IsSubArr(T &,int,T &,int)' : template parameter 'T' is ambiguous
please help to resolve this issue ?

Answer 1

Since array[a] and array[b] where a != b are 2 different types, you'll need 2 type templates args.
A work around would be to use pointers.
+ template <class T> int IsSubArr(T* a, int a_len, T* b, int b_len)
+ T* s = a; T*l = b;

Answer 2

You defined the first and the third parameters as references
template <class T> int IsSubArr(T& a, int a_len, T& b, int b_len)
^^^^ ^^^^
and pass as arguments for these parameters two arrays with different types
int array[5] = {9,4,6,2,1};
int sub_arr[3] = {6,2,1};
//...
index = IsSubArr(array,alen,sub_arr,slen);
^^^^^ ^^^^^^^
The first argument has type int[5] and the third argument has type int[3]
So the compiler is unable to deduce the referenced type T.
If you are going to use arrays with the function then you could declare it like
template <class T, size_t N1, size_t N2>
int IsSubArr( T ( &a )[N1], T ( &b )[N2] );
Or you could use pointers instead of the references to arrays
template <class T> int IsSubArr( T *a, size_t a_len, T *b, size_t b_len );
Take into account that this declaration within the function
T& s=a,l=b;
is also wrong. It is equivalent to the following declarations
T& s=a;
T l=b;
That is the first declaration declares a reference to an array while the second declaration declares an array and tries to initialize it with another array. However arrays do not have a copy constructor and the compiler will issue one more error. And you may not reassign a reference.
You should know that there is standard algorithm std::search declared in header <algorithm> that can do the job you want to do with your function.

Answer 3

It's because array and sub_arr are two different types. array is of type int[5] while sub_arr is of type int[3]. The array dimensions are part of the type.
Either change the function to use two different templates arguments, one for each array, or use pointer (e.g. T*)
There's also another error, that you will continue to have if you keep using arrays and two different template arguments, and that is that you can't change references. Once you have assigned to the variable s and l in the function, those can't be made to reference anything else. The latter assignments of the s and l variables will fail because there you try to assign the arrays to each other, which you can not do, you can only copy arrays.
If you use pointers instead, then this won't be a problem.