Why is C++ casting the string literal I pass in as a bool rather than a string?
#include
using namespace std;
class A
{
public:
If you have C++11 you can use a delegating constructor:
A(char const* s) : A(std::string(s)) { }
The reason the boolean converting-constructor is chosen over the one for std::string
is because the conversion from char const*
to bool
is a standard conversion while the one to std::string
is a user-defined conversion. Standard conversions have a greater rank than user-defined conversions.
When selecting an overloaded method this is the order that the compiler tries:
A pointer doesn't promote to bool
but it does convert to bool
. A char*
uses an std
operator to convert to std::string
. Note that if char*
used the same number in this list to convert to both bool
and std::string
it would be ambiguous which method the compiler should choose, so the compiler would throw an "ambiguous overload" error.
I would throw my weight behind 0x499602D2's solution. If you have C++11 your best bet is to call: A(char* s) : A(std::string(s)){}
If you don't have C++11 then I would create an A(char* s)
constructor and abstract the logic of the A(std::string)
constructor into a method and call that method from both constructors.
http://www.learncpp.com/cpp-tutorial/76-function-overloading/
Recently I passed this problem too, let me share another way.
You can change the bool
constructor to a unsigned char
. So the decay and implict conversion of string literal don't happen, and the std::string
constructor takes place.
class A
{
public:
A(string v)
{
cout << v;
}
A(unsigned char v)
{
cout << static_cast<bool>(v);
}
};
int main()
{
A("Hello"); // <- Call A(string)
A(false); // <- Call A(unsigned char)
}
This way you don't need to provide always overloads to std::string
and const char*
neither making code bloat constructing the std::string
at the client call site.
I don't claim that's better, but it's simpler.
With
A(string("hello"));
it will give the expected result.
Why is it so ?
It's because of the standard conversions:
const char*
pointerbool
(section 4.12 of the standard: "A prvalue of (...) pointer (...) type can be converted to a prvalue of type bool." )string
is not considered because section 12.3 of standard explains that "Type conversions of class objects can be specified by constructors and by conversion functions. These conversions are called user-defined conversions" and "User-defined conversions are applied only where they are unambiguous". Wouldn't you have the bool
constructor the std::string
conversion would be done implicitely. How to get what you expected ?
Just add the missing constructor for string litterals:
A(const char* v)
{
cout << v; // or convert v to string if you want to store it in a string member
}
Of course, instead of rewriting a constructor from scratch, you could opt for a delegate as suggested by 0x499602D2 in another answer.