The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.
Here's an example class with a constructor that can be used for implicit conversions:
class Foo
{
public:
// single parameter constructor, can be used as an implicit conversion
Foo (int foo) : m_foo (foo)
{
}
int GetFoo () { return m_foo; }
private:
int m_foo;
};
Here's a simple function that takes a Foo
object:
void DoBar (Foo foo)
{
int i = foo.GetFoo ();
}
and here's where the DoBar
function is called:
int main ()
{
DoBar (42);
}
The argument is not a Foo
object, but an int
. However, there exists a constructor for Foo
that takes an int
so this constructor can be used to convert the parameter to the correct type.
The compiler is allowed to do this once for each parameter.
Prefixing the explicit
keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call DoBar (42)
. It is now necessary to call for conversion explicitly with DoBar (Foo (42))
The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:
- You have a
MyString
class with a constructor that constructs a string of the given size. You have a function print(const MyString&)
(as well as an overload print (char *string)
), and you call print(3)
(when you actually intended to call print("3")
). You expect it to print "3", but it prints an empty string of length 3 instead.
Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.
Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example:
template<typename T>
struct Foo
{
T bar;
void doSomething(T param) {/* do stuff using T */}
};
// somewhere in a .cpp
Foo<int> f;
When reading this line, the compiler will create a new class (let's call it FooInt
), which is equivalent to the following:
struct FooInt
{
int bar;
void doSomething(int param) {/* do stuff using int */}
}
Consequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int
). If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.
A common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example .tpp), and include this implementation file at the end of the header.
Foo.h
template <typename T>
struct Foo
{
void doSomething(T param);
};
#include "Foo.tpp"
Foo.tpp
template <typename T>
void Foo<T>::doSomething(T param)
{
//implementation
}
This way, implementation is still separated from declaration, but is accessible to the compiler.
Alternative solution
Another solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:
Foo.h
// no implementation
template <typename T> struct Foo { ... };
Foo.cpp
// implementation of Foo's methods
// explicit instantiations
template class Foo<int>;
template class Foo<float>;
// You will only be able to use Foo with int or float
If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject.
Best Answer
auto
was a keyword that C++ "inherited" from C that had been there nearly forever, but virtually never used because there were only two possible conditions: either it wasn't allowed, or else it was assumed by default.The use of
auto
to mean a deduced type was new with C++11.At the same time,
auto x = initializer
deduces the type ofx
from the type ofinitializer
the same way as template type deduction works for function templates. Consider a function template like this:At point A, a type has been assigned to
T
based on the value passed for the parameter towhatever
. When you doauto x = initializer;
, the same type deduction is used to determine the type forx
from the type ofinitializer
that's used to initialize it.This means that most of the type deduction mechanics a compiler needs to implement
auto
were already present and used for templates on any compiler that even sort of attempted to implement C++98/03. As such, adding support forauto
was apparently fairly easy for essentially all the compiler teams--it was added quite quickly, and there seem to have been few bugs related to it either.When this answer was originally written (in 2011, before the ink was dry on the C++ 11 standard)
auto
was already quite portable. Nowadays, it's thoroughly portable among all the mainstream compilers. The only obvious reasons to avoid it would be if you need to write code that's compatible with a C compiler, or you have a specific need to target some niche compiler that you know doesn't support it (e.g., a few people still write code for MS-DOS using compilers from Borland, Watcom, etc., that haven't seen significant upgrades in decades). If you're using a reasonably current version of any of the mainstream compilers, there's no reason to avoid it at all though.More recent revisions of the standard have added a few new places that
auto
can be used. Starting with C++14, you can useauto
for the type of a parameter to a lambda:This does essentially the same thing as the example above--even though it doesn't explicitly use
template
syntax, this is basically a template that deduces the type of the parameter, and instantiates the template over that type.That was convenient and useful enough that in C++20, the same capability was added for normal functions, not just lambdas.
But, just as before all of this really comes down to using the same basic type deduction mechanism as we've had for function templates since C++98.
auto
allows that to be used in more places, and more conveniently, but the underlying heavy lifting remains the same.