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Let's assume we have a template function "foo":
template<class T>
void foo(T arg)
{ ... }
I can make specialization for some particular type, e.g.
template<>
void foo(int arg)
{ ... }
If I wanted to use the same specialization for all builtin numeric types (int, float, double etc.) I would write those lines many times. I know that body can be thrown out to another function and just call of this is to be made in every specialization's body, however it would be nicer if i could avoid writting this "void foo(..." for every type. Is there any possibility to tell the compiler that I want to use this specialization for all this types?
746
The act of creating a new definition of a function, class, or member of a class from a template declaration and one or more template arguments is called template instantiation. The definition created from a template instantiation is called a specialization.
Member templates that are classes are referred to as nested class templates. Member templates that are functions are discussed in Member Function Templates. Nested class templates are declared as class templates inside the scope of the outer class. They can be defined inside or outside of the enclosing class.
Function Templates with Multiple ParametersYou can also use multiple parameters in your function template. The above syntax will accept any number of arguments of different types. Above, we used two generic types such as A and B in the template function.
A template is not a class or a function.
You could use std::numeric_limits to see whether a type is a numeric type (is_specialized is true for all float and integer fundamental types).
// small utility
template<bool> struct bool2type { };
// numeric
template<typename T>
void fooImpl(T arg, bool2type<true>) {
}
// not numeric
template<typename T>
void fooImpl(T arg, bool2type<false>) {
}
template<class T>
void foo(T arg)
{ fooImpl(arg, bool2type<std::numeric_limits<T>::is_specialized>()); }
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