When overloading a method, I believe the compiler will choose the simpler match when multiple matches are available.
Consider this code:
#include <iostream>
#include <string>
struct A {
static void foo(const char *str) {
std::cout << "1: " << str << std::endl;
}
template<int N> static void foo(const char (&str)[N]) {
std::cout << "2: " << str << std::endl;
}
};
int main()
{
A::foo("hello");
}
The output is 1: hello
. Yet, if I comment out the static void foo(const char *str)
method, it compiles fine and outputs 2: hello
.
How can I have both methods on a class such that arrays with known size will call the template method, and pointer types call the non-template method?
I tried the following:
struct A {
template<class _Ty = char>
static void foo(const _Ty *str) {
std::cout << "1: " << str << std::endl;
}
template<int N> static void foo(const char (&str)[N]) {
std::cout << "2: " << str << std::endl;
}
};
But g++ gives me the following error:
In function 'int main()':
17:17: error: call of overloaded 'foo(const char [6])' is ambiguous
17:17: note: candidates are:
6:15: note: static void A::foo(const _Ty*) [with _Ty = char]
10:32: note: static void A::foo(const char (&)[N]) [with int N = 6]
As suggested by T.C., this works:
struct A {
template<class T, typename = typename std::enable_if<std::is_same<T, char>::value>::type>
static void foo(const T * const & str) {
std::cout << "1: " << str << std::endl;
}
template<int N> static void foo(const char (&str)[N]) {
std::cout << "2: " << str << std::endl;
}
};
int main()
{
A::foo("hello1");
const char *c = "hello2";
A::foo(c);
char *c2 = new char[7];
::strcpy(c2, "hello3");
A::foo(c2);
// does not compile
// int *c3;
// A::foo(c3);
}
Outputs:
2: hello1
1: hello2
1: hello3
I wish I did not have to template the pointer method since it opens the door to misuses with unexpected types, but I can live with this solution.
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