cin >> function template arg of unknown type

Question

I have a function template and main as follows:

 template <class type > type* calculate(type inputVal) {
       type val;
       static int counter = 0;
       static type sum = inputVal;
       static type average = inputVal;
       static type* address = &sum

       do {
          cout << "Enter value: ";
          cin >> val;
          counter++;
          sum += val;
          average = sum / counter;
       } while (!cin.eof());
      return address;
 }

void main() {
      int num;
      cout << "Enter Value: ";
      cin >> num;
      int *ptr = calculate(num);
      cout << "SUM: " << *ptr << " AVG: " << *(ptr+1);
 }

My problem is that this should work with different input types instead of just int, so if a user first enters a float it'll treat everything as that type, as wells as if the user enters a char.

Also the template function cannot print the end values.

1. Declareing or defining `void main()` in global namespace is illegal in standard C++. You should use `int main()` as entry point. 2. Semicolons are missing after `static type* address = &sum`, `cin >> num` and `calculate(num)` 3. `!cin.eof()` as loop condition [seems bad](http://stackoverflow.com/questions/5605125/why-is-iostreameof-inside-a-loop-condition-considered-wrong) and I suggest you should check if `cin >> val` is successful instead. — MikeCAT, Jul 14 '16 at 05:16
It's not possible. A program cannot decide what type a variable will have based on what it reads from the standard input. All types of all variables must be known in advance. — n. m. could be an AI, Jul 14 '16 at 05:45

MikeCAT · Accepted Answer · 2016-07-14T05:33:45.297

A normal variable sum is treated as one-element array for pointer aritimetic (N3337 5.7 Additive operators) and when ptr points at it, ptr+1 doesn't point at a valid object and therefore must not be dereferenced. If you want continuous memory region, use arrays.

Also note that

Checking !cin.eof() after updating sum and average doesn't seem a good idea because it will use an invalid (duplicate) data. Check if input is successful before processing the data.
Declareing void main() (or main with return type other than int) in the global namespace is illegal in the standard C++. Unless you have some special reason -- for example, your boss or teacher forbids writing standard-compliant code -- you should use int main() (in this case).
You should initialize counter to 1 to put inputVal in the number. Avoiding getting input as argument to avoid writing duplicate code seems better.

Try this:

#include <iostream>
using std::cin;
using std::cout;

template <class type > type* calculate(type inputVal) {
  type val;
  static int counter = 1;
  static type buffer[2];
  type& sum=buffer[0];
  type& average=buffer[1];
  sum=average=inputVal;
  static type* address = buffer;

  for(;;) {
    cout << "Enter value: ";
    if(!(cin >> val)) break;
    counter++;
    sum += val;
    average = sum / counter;
  }
  return address;
}

int main() {
  int num;
  cout << "Enter Value: ";
  cin >> num;
  int *ptr = calculate(num);
  cout << "SUM: " << *ptr << " AVG: " << *(ptr+1);
}

Or this without input from the argument:

#include <iostream>
using std::cin;
using std::cout;

template <class type > type* calculate() {
  type val;
  static int counter = 0;
  static type buffer[2];
  type& sum=buffer[0];
  type& average=buffer[1];
  sum=0; // the type used has to be capable to be converted from integer 0
  average = 0;
  static type* address = buffer;

  for(;;) {
    cout << "Enter value: ";
    if(!(cin >> val)) break;
    counter++;
    sum += val;
    average = sum / counter;
  }
  return address;
}

int main() {
  int *ptr = calculate<int>();
  cout << "SUM: " << *ptr << " AVG: " << *(ptr+1);
}

is there anyway to do it without the use of arrays? and also what if the user wanted to calculate a float value? — leakybytes, Jul 14 '16 at 05:32
@GeorgeVentura Use structures to avoid using arrays in this case. Let the user get code to deal with `float` if the user want it. — MikeCAT, Jul 14 '16 at 05:34
I tried this program is not terminating can you tell when this condition if(!(cin >> val)) will be true — PapaDiHatti, Jul 14 '16 at 05:35
@GeorgeVentura When the input fails and `cin >> val` returns `false`. — MikeCAT, Jul 14 '16 at 05:36
Also this program will give wrong output for average because there is one input already in main so counter should be initialized to 1 — PapaDiHatti, Jul 14 '16 at 05:38
@Kapil In the first code, `counter` is already initialized to 1. In the second code, I don't think there is one input already in main. — MikeCAT, Jul 14 '16 at 05:40
sorry i just refresh my page. Also one question related to this, Is there any way by which template can deduce its type from left hand side i.e if we call int* ptr = calculate() then it take its type as int — PapaDiHatti, Jul 14 '16 at 05:48

cin >> function template arg of unknown type

1 Answers1