Instead of explaining the behavior of compilers, I'll try to explain what the standard says.
Primary Example
To direct-initialize b1
from {{{"test"}}}
, overload resolution applies to choose the best constructor of B
. Because there is no implicit conversion from {{{"test"}}}
to B&
(list initializer is not a lvalue), the constructor B(B&)
is not viable. We then focus on the constructor B(A)
, and check whether it is viable.
To determine the implicit conversion sequence from {{{"test"}}}
to A
(I will use the notation {{{"test"}}}
-> A
for simplicity), overload resolution applies to choose the best constructor of A
, so we need to compare {{"test"}}
-> const char*
and {{"test"}}
-> std::string
(note the outermost layer of braces is elided) according to [over.match.list]/1:
When objects of non-aggregate class type T are list-initialized such that [dcl.init.list] specifies that overload resolution is performed according to the rules in this subclause, overload resolution selects the constructor in two phases:
Initially, the candidate functions are the initializer-list constructors ([dcl.init.list]) of the class T...
If no viable initializer-list constructor is found, overload resolution is performed again, where the candidate functions are all the constructors of the class T and the argument list consists of the elements of the initializer list.
... In copy-list-initialization, if an explicit constructor is chosen, the initialization is ill-formed.
Note all constructors are considered here regardless of the specifier explicit
.
{{"test"}}
-> const char*
does not exist according to [over.ics.list]/10 and [over.ics.list]/11:
Otherwise, if the parameter type is not a class:
In all cases other than those enumerated above, no conversion is possible.
To determine {{"test"}}
-> std::string
, the same process is taken, and overload resolution chooses the constructor of std::string
that takes a parameter of type const char*
.
As a result, {{{"test"}}}
-> A
is done by choosing the constructor A(std::string)
.
Variations
What if explicit
is removed?
The process does not change. GCC will choose the constructor A(const char*)
while Clang will choose the constructor A(std::string)
. I think it is a bug for GCC.
What if there are only two layers of braces in the initializer of b1
?
Note {{"test"}}
-> const char*
does not exist but {"test"}
-> const char*
exists. So if there are only two layers of braces in the initializer of b1
, the constructor A(const char*)
is chosen because {"test"}
-> const char*
is better than {"test"}
-> std::string
. As a result, an explicit constructor is chosen in copy-list-initialization (initialization of the parameter A
in the constructor B(A)
from {"test"}
), then the program is ill-formed.
What if the constructor B(const B&)
is declared?
Note this also happens if the declaration of B(B&)
is removed. This time we need to compare {{{"test"}}}
-> A
and {{{"test"}}}
-> const B&
, or {{{"test"}}}
-> const B
equivalently.
To determine {{{"test"}}}
-> const B
, the process described above is taken. We need to compare {{"test"}}
-> A
and {{"test"}}
-> const B&
. Note {{"test"}}
-> const B&
does not exist according to [over.best.ics]/4:
However, if the target is
— the first parameter of a constructor or
— the implicit object parameter of a user-defined conversion function
and the constructor or user-defined conversion function is a candidate by
— [over.match.ctor], when the argument is the temporary in the second step of a class copy-initialization,
— [over.match.copy], [over.match.conv], or [over.match.ref] (in all cases), or
— the second phase of [over.match.list] when the initializer list has exactly one element that is itself an initializer list, and the target is the first parameter of a constructor of class X, and the conversion is to X or
reference to cv X,
user-defined conversion sequences are not considered.
To determine {{"test"}}
-> A
, the process described above is taken again. This is almost the same as the case we talked in the previous subsection. As a result, the constructor A(const char*)
is chosen. Note the constructor is chosen here to determine {{{"test"}}}
-> const B
, and does not apply actually. This is permitted though the constructor is explicit.
As a result, {{{"test"}}}
-> const B
is done by choosing the constructor B(A)
, then the constructor A(const char*)
. Now both {{{"test"}}}
-> A
and {{{"test"}}}
-> const B
are user-defined conversion sequences and neither is better than the other, so the initialization of b1
is ambiguous.
What if the parentheses is replaced by braces?
According to [over.best.ics]/4, which is block-quoted in the previous subsection, the user defined conversion {{{"test"}}}
-> const B&
is not considered. So the result is the same as the primary example even if the constructor B(const B&)
is declared.