如何在c ++中使用泛型函数？

Question

I have written this code in which if I uncomment the 2nd last line I get error - "template argument deduction/substitution failed: ". Is it because of some limit to generic functions in C++? Also my program doesn't print floating answer for the array b . Is there anything I can do for that? (sorry for asking 2 questions in single post.)
PS: I have just started learning C++.

#include <iostream>
 using namespace std;

 template <class T>
 T sumArray(  T arr[], int size, T s =0)
 {
     int i;
     for(i=0;i<size;i++)
     {  s += arr[i];
     }
     return s;
 }

 int main()
 {
     int a[] = {1,2,3};
     double b[] = {1.0,2.0,3.0};
     cout << sumArray(a,3) << endl;
     cout << sumArray(b,3) << endl;
     cout << sumArray(a,3,10) << endl;
     //cout << sumArray(b,3,40) << endl; //uncommenting this line gives error

     return 0;
 }

EDIT 1: After changing 40 to 40.0, the code works. Here is the output I get:
6
6
16
46
I still don't get the floating answer in 2nd case. Any suggestion ?

Answer 1

The reason is that compiler can not deduce the type for T .

How it should understand what T is for your last example? The type of the first argument ( b ) is double[] , while it is T[] in the function definition. Therefore it looks like that T should be double . However, the type of the third argument ( 40 ) is int , so it looks like T should be int . Hence the error.

Changing 40 to 40.0 makes it work. Another approach is to use two different types in template declaration:

#include <iostream>
 using namespace std;

 template <class T, class S = T>
 T sumArray(  T arr[], int size, S s =0)
 {
     int i;
     T res = s;
     for(i=0;i<size;i++)
     {  res += arr[i];
     }
     return res;
 }

 int main()
 {
     int a[] = {1,2,3};
     double b[] = {1.0,2.0,3.1};
     cout << sumArray(a,3) << endl;
     cout << sumArray(b,3) << endl;
     cout << sumArray(a,3,10) << endl;
     cout << sumArray(b,3,40) << endl; //uncommenting this line gives error

     return 0;
 }

Note that I had to cast s to T explicitly, otherwise the last example will lose fractional part.

However, this solution will still not work for sumArray(a,3,10.1) because it will cast 10.1 to int , so if this is also a possible use case, a more accurate treatment is required. A fully working example using c++11 features might be like

 template <class T, class S = T>
 auto sumArray(T arr[], int size, S s=0) -> decltype(s+arr[0])
 {
    int i;
    decltype(s+arr[0]) res = s;
    ...

Another possible improvement for this template function is auto-deduction of array size, see TartanLlama's answer.

Answer 2

sumArray(b,3,40)

The type of 40 is int , but the type of b is double[3] . When you pass these in as arguments, the compiler gets conflicting types for T .

A simple way to fix this is to just pass in a double :

sumArray(b,3,40.0)

However, you would probably be better off allowing conversions at the call site by adding another template parameter. You can also add one to deduce the size of the array for you so that you don't need to pass it explicitly:

template <class T, class U=T, std::size_t size>
U sumArray(T (&arr) [size], U s = 0)

The U parameter is defaulted to T to support the default value for s . Note that to deduce the size of the array, we need to pass a reference to it rather than passing by value, which would result in it decaying to a pointer.

Calling now looks like this:

sumArray(b,40)

Live Demo

Answer 3

In

template <class T>
T sumArray(  T arr[], int size, T s =0)
             ^                  ^

Both (deducible) T should match.

In sumArray(b, 3, 40) , it is double for the first one, and int for the second one.

There is several possibilities to fix problem

• at the call site, call sumArray(b, 3, 40.0) or sumArray<double>(b, 3, 40);

• Use extra parameter:

   template <typename T, typename S> auto sumArray(T arr[], int size, S s = 0) 

  Return type may be T , S , or decltype(arr[0] + s) depending of your needs.

• make a parameter non deducible:

   template <typename T> struct identity { using type = T;}; // or template<typename T> using identity = std::enable_if<true, T>; template <typename T> T sumArray(T arr[], int size, typename identity<T>::type s = 0) 

Answer 4

Should be

sumArray(b,3,40.0)

so, T will be deduced to double . In your code it's int .

Answer 5

演绎失败时的另一个选择是明确告诉编译器你的意思：

cout << sumArray<double>(b,3,40) << endl;

Answer 6

The compiler does not know whether T should be int or double .

You might want to do the following, in order to preserve the highest precision of the types passed:

template <class T, class S>
std::common_type_t <T, S> sumArray (T arr [], std::size_t size, S s = 0)
{
   std::common_type_t <T, S> sum = s;
   for (std::size_t i = 0; i != size; ++i)
   {
      sum += arr[i];
   }
   return sum;
}

The function you are writing, however, already exists. It's std::accumulate :

std::cout << std::accumulate (std::begin (b), std::end (b), 0.0) << std::endl;

Answer 7

Templates only accept one type of data, for example if you send an array of double, then the runtime will deduce :

Template = double[]

so every time he will see it he will expect an array of doubles.

sumArray(b,3,40) passes "b" (which is an array of doubles) but then you pass "40" which the runtime cannot implicitly convert to double.

So the code

sumArray(b,3,40.0)

will work