Implicit type conversion in c++ template

Question

I have a function template:

template<typename T>
void fun(T a, T b){
         .......
}

int a = 0;
double b = 1.2;
f(a, b);

can a be converted to double automatically?

Why was the question downvoted? It is a legitimate question. — SergeyA, Apr 15 '16 at 13:57
@SergeyA I did not downvote (reading it only now), but the question is a bit ambiguous and doesn't explain rationale at all. It doesn't explain whether `f(b, a)` should convert `double` to `int` or `int` to `double` or fail to compile. It doesn't explain whether he wants this treatment also for `short a; f(a, b)`. Neither which are the supported types of `fun`. I would have answered "*yes, use fun(double a, T b)*", but the question has no information on whether that's useless or not. Not enough reason to downvote for me. I will just move on and won't be bothered. — Johannes Schaub - litb, Apr 15 '16 at 15:32

songyuanyao · Answer 1 · 2016-04-15T14:01:19.817

can a be converted to double automatically?

No, because it's ambiguous between fun<int> and fun<double>, when deducing the type of T in template argument deduction.

You could specify the template argument explicitly, to make a implicitly converted to double:

int a = 0;
double b = 1.2;
fun<double>(a, b);

or add an explicit conversion, to make the template argument deduction unambiguous:

int a = 0;
double b = 1.2;
fun(static_cast<double>(a), b);

score 0 · Answer 2 · answered Apr 15 '16 at 13:56

No it can't. There are no conversions done during the template deduction process. In this case, we deduce T independently from a and b, getting int for a and double for b - since we deduced T to be two different types, that is a deduction failure.

If you want to do conversions, the simplest thing would either be to explicitly do it yourself:

f(static_cast<double>(a), b);

Or to explicitly provide the template parameter to f so that no deduction happens:

f<double>(a, b);

score 0 · Answer 3 · answered Apr 15 '16 at 16:02

If your intention is that parameter a be converted to type of parameter b, then the following template can be used instead of yours:

template<typename Ta, typename T>
void fun(Ta aTa, T b) {
    T& a = static_cast<T>(aTa);
    /* ... a and b have the same type T ... */
}

int a = 0;
double b = 1.2;
fun(a, b);    // works fine

Implicit type conversion in c++ template

3 Answers3

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