If I have a function template with typename T
, where the compiler can set the type by itself, I do not have to write the type explicitly when I call the function like:
template < typename T >
T min( T v1, T v2 ) {
return ( v1 < v2 ) ? v1: v2;
}
int i1 = 1, i2 = 2; int i3 = min( i1, i2 ); //no explicit <type>
But if I have a function template with two different typenames like:
template < typename TOut, typename TIn >
TOut round( TIn v ) {
return (TOut)( v + 0.5 );
}
double d = 1.54;
int i = round<int>(d); //explicit <int>
Is it true that I always have to specify at least 1 typename? I assume the reason is because C++ can not distinguish functions between different return types.
But if I use a void function and handover a reference, again I must not explicitly specify the return typename:
template < typename TOut, typename TIn >
void round( TOut & vret, TIn vin ) {
vret = (TOut)(vin + 0.5);
}
double d = 1.54;
int i; round(i, d); //no explicit <int>
Should the conclusion be to avoid functions with return and more prefer void
functions that return via a reference when writing templates? Or is there a possibility to avoid explicitly writing the return type? Something like "type inference" for templates. Is "type inference" possible in C++0x?
Defining a Function TemplateA function template starts with the keyword template followed by template parameter(s) inside <> which is followed by the function definition. In the above code, T is a template argument that accepts different data types ( int , float , etc.), and typename is a keyword.
A non-template class can have template member functions, if required. Notice the syntax. Unlike a member function for a template class, a template member function is just like a free template function but scoped to its containing class.
If we want to define a different implementation for a template when a specific type is passed as template parameter, we can declare a specialization of that template.
Yes, you need the inline specifier there. The ODR (one-definition rule) states that there must be exactly one definition of a variable, function, class, enum or template.
Overload resolution is done only based on function arguments; the return value is not used at all. If the return type cannot be determined based on the arguments, you will have to specify it explicitly.
I would not go down the path of "returning" a value through a reference parameter; that makes the calling code unclear. For example, I'd prefer this:
double x = round<double>(y);
over this:
double x;
round(x, y);
because in the latter case, it's easy to confuse input and output, and it's not at all clear that x
is being modified.
In the particular case of round
, you probably need only one or two types for TOut
anyway, so you could just leave that template argument out:
template<typename TIn>
int roundToInt(TIn v) {
return (int)(v + 0.5);
}
I find roundToInt(x)
a little clearer than round<int>(x)
because it's clear what the int
type is used for.
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