I can't figure out why the following code compiles fine:
#include <iostream>
void bar(int x) {
std::cout << "int " << x << std::endl;
}
void bar(double x) {
std::cout << "double " << x << std::endl;
}
template <typename A, typename B> // Note the order of A and B.
void foo(B x) {
bar((A)x);
}
int main() {
int x = 1;
double y = 2;
foo<int>(x); // Compiles OK.
foo<double>(y); // Compiles OK.
return 0;
}
But if I switch the order of A
and B
as below, then it won't compile:
#include <iostream>
void bar(int x) {
std::cout << "int " << x << std::endl;
}
void bar(double x) {
std::cout << "double " << x << std::endl;
}
template <typename B, typename A> // Order of A and B are switched.
void foo(B x) {
bar((A)x);
}
int main() {
int x = 1;
double y = 2;
foo<int>(x); // error: no matching function for call to ‘foo(int&)’
foo<double>(y); // error: no matching function for call to ‘foo(double&)’
return 0;
}
EDIT: Ad-hoc explanations are welcome, but would be better if someone can point out exact what the spec. says. Thanks!
Template classes and functions can make use of another kind of template parameter known as a non-type parameter. A template non-type parameter is a template parameter where the type of the parameter is predefined and is substituted for a constexpr value passed in as an argument.
To instantiate a template function explicitly, follow the template keyword by a declaration (not definition) for the function, with the function identifier followed by the template arguments. template float twice<float>(float original); Template arguments may be omitted when the compiler can infer them.
Unless a template specialization has been explicitly instantiated or explicitly specialized, the compiler will generate a specialization for the template only when it needs the definition. This is called implicit instantiation.
When a function template is first called for each type, the compiler creates an instantiation. Each instantiation is a version of the templated function specialized for the type. This instantiation will be called every time the function is used for the type.
In the first one, the compiler knows that A
is int
because you specifically tell it so with foo<int>
, and it knows that B
is also int
because of the parameter that you pass it. So both A
and B
are known or can be deduced (you could say: A
is supplied, B
is implied).
However, in the second one, since B
comes first and A
doesn't appear in the parameter list, the compiler can't tell what A
is and gives you an error. You're explicitly telling it what B
is with foo<int>
, and then the parameter you pass is also a B
which, at the call, is an int
which agrees with your previous explicit definition of B
, but no mention is made of A
, implicitly or explicitly, so the compiler must stop and error.
You don't really need the standard for this, it's just common sense. What on earth would A
be in the second one?
Thanks for asking this question though, because I didn't realise you could explicitly specify some parameters and implicitly specify others in the parameter list before this.
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