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is there any way to force functions to only take in vectors of integers (int, unsigned int, uint32_t, etc.) and only those? im trying to write a simple function that returns the sum of all the values with the same return type (since i cant be sure whether or not the value will be greater than (2^32 - 1). however, since std::vector<T> doesnt have a cout operator for the entire type, i cannot do sum(vector<vector<T> >), since it will return a vector (ignoring the fact that vector + vector doesnt work). i do not wish to overload every type to cout something because i wont be needing it. i just want the function to work when T is some form of an int (and float if possible)

ive tried using try/except, but codeblocks catches the operators of the types, so i cant compile if i do a sum(vector <vector <T> >)

template <typename T>
T sum(std::vector <T> in){
    try{
        T total = 0;
        for(int x = 0; x < in.size(); x++)
            total += in[x];
        return total;
    }
    catch (int e){
        std::cout << "Error " << e << " has occured." << std::endl;
        exit(e);
    }
}
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4
  • if T is an integral type, it is impossible that an int is thrown in your code. Commented Jun 9, 2011 at 4:52
  • im not familiar with using try/catch yet Commented Jun 9, 2011 at 5:11
  • 2
    I suggest to never use them then until you know what they're for. As a rule of thumb, the fewer there are in your code, the better. Commented Jun 9, 2011 at 5:16
  • 1
    Just FYI - if your sum() template is instantiated for a type for which T total = 0 is invalid, then you'll get a compiler error. The situation is never allowed to linger and manifest at run-time, which is when exceptions and try/catch blocks execute. C++ doesn't do compilation at run-time. Separately, taking your input vector by const reference prevents a temporary copy of the entire container being generated every time the function is called... Gene illustrates this but doesn't explain or justify it. Commented Jun 9, 2011 at 5:19

2 Answers 2

Reset to default
3 
template <typename T>
typename std::enable_if<
    std::is_integral<typename std::remove_reference<T>::type>::value,
    T>::type
sum(const std::vector<T>& in) { return std::accumulate(in.begin(), in.end(), 0); }
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2 Comments

Can't have an std::vector of references (not that it makes your code invalid).
@Luc: correct, after typing those things again and again it became almost automatic :)
2

SFINAE to the rescue.

#include <type_traits>
//#include <tr1/type_traits> // for C++03, use std::tr1::

template<bool, class T = void>
struct enable_if{};

template<class T>
struct enable_if<true,T>{
  typedef T type;
};

template<class T>
typename enable_if<
    std::is_arithmetic<T>::value,
    T
>::type sum(std::vector <T> in){
    T total = 0;
    for(int x = 0; x < in.size(); x++)
        total += in[x];
    return total;
}
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7 Comments

both solutions give me errors with error: 'enable_if' in namespace 'std' does not name a type
@calccrypto Did you include <type_traits> as it appears in Xeo's answer? Is your compiler C++0x aware?
yes. i tried both. i also enabled c++0x and its sending me to crazy places saying that there are errors (probably due to the new rules)
calccrypto: hard to diagnose the C++0x issues without proper details of the code and error message. Avoiding C++0x, though probably most hassle than it's worth if you're not already using boost, boost also provides a C++03 compatible enable_if template and is_arithmetic<> etc. tests you could use....
if you want to enable both integral and floating point you can just use std::is_arithmetic<T>
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