Question

Example: Assuming f to be a template function, having two arguments:

f (1, 2) In this call, does the template function assume that its arguments are int, or short, or anything else?

EDIT 1:

The template function declaration:

template <typename dataTypeA, typename dataTypeB> dataTypeB functionX (dataTypeA argA, dataTypeB argB)

Answer 1

As @David already said, as far as your question is concerned, there’s no such thing as “making assumptions”. The literals simply have types, which are the types that a function template may use for type deduction. Remember that conversions considered as part of the template matching, though!

So, let’s say you have this function template:

template <typename T> void foo(T x, T y);

Then if you call foo(1, 2), this will be called with T = int.

If you say foo(1u, 2u), the deduction is T = unsigned int.

If you say anything mixed like foo(1u, 2), there is no preferred match and the compiler will report an error.

Since there is no short literal in C or C++, if you want to explicitly call the function foo<short>, you can either say so, or create temporary explicitshort arguments:

foo<short int>(3, 4);
foo<short int>(3u, 4l);  // also OK because of conversion
foo(short(3), short(4)); // deduction

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Update: In light of your edit, note that since you’re only matching one argument per template parameter, you won’t have any trouble with ambiguous matching.