In the following piece of code (live on coliru):
#include <iostream>
#include <string>
int main()
{
struct S {
operator bool () const { return false; }
operator std::string () const { return "false"; }
} s;
std::cout << s << "\n"; // outputs 0
}
How does the compiler choose to pick the implicit conversion to bool
over std::string
?
My hypothesis is that in this case, it might be purely the order of declaration of the different flavours of std::basic_ostream::operator<<
, but is it all? Does the standard say something about picking a specific implicit conversion?
Recall that std::string
is not a standalone type, it's really a class template specialization - std::basic_string<char>
. The very important detail is that the potential overload for streaming a std::string
does not take a std::string const&
argument, it is a function template that deduces a std::basic_string const&
:
template <class CharT, class Traits, class Allocator>
std::basic_ostream<CharT, Traits>&
operator<<(std::basic_ostream<CharT, Traits>& os,
const std::basic_string<CharT, Traits, Allocator>& str);
Template deduction never considers conversions. Name lookup will find this function template, and then discard at as being non-viable due to deduction failure. S
is not a basic_string<CharT, Traits, Allocator>
for any such types, so we're done. The only viable stream operators would be all the integral ones, of which bool
is the best match.
If there specifically was a function with signature:
std::ostream& operator<<(std::ostream&, std::string const& );
Then the call would be ambiguous - you'd get two user-defined conversions that would be equivalently ranked.
This is easy to verify by using our own functions instead of the million overloads for operator<<
:
void foo(bool ); // #1
void foo(std::string ); // #2
void bar(bool ); // #3
template <class C, class T, class A>
void bar(std::basic_string<C,T,A> ); // #4
foo(S{}); // error: ambiguous
bar(S{}); // calls #3