Implicit Template Parameters

Question

The following code generates a compile error in Xcode:

template 
struct Foo
{
    Foo(T Value)
    {
    }
};

int main()
{
    Foo MyFoo(123);         


        

Answer 1

compiler can figure out template parameter type only for templated functions, not for classes/structs

Answer 2

In C++11 you can use decltype:

int myint = 123;
Foo<decltype(myint)> MyFoo(myint);

Answer 3

It's not a bug, it's non-existing feature. You have to fully specify class/structure template arguments during instantiation, always, the types are not inferred as they can be for function templates.

Answer 4

Compiler can deduce the template argument such case:

template<typename T>
void fun(T param)
{
    //code...
}

fun(100);    //T is deduced as int;
fun(100.0);  //T is deduced as double
fun(100.0f); //T is deduced as float

Foo<int> foo(100);
fun(foo);    //T is deduced as Foo<int>;

Foo<char> bar('A');
fun(bar);    //T is deduced as Foo<char>;

Actually template argument deduction is a huge topic. Read this article at ACCU:

The C++ Template Argument Deduction

Answer 5

The constructor could in theory infer the type of the object it is constructing, but the statement:

Foo MyFoo(123);

Is allocating temporary space for MyFoo and must know the fully-qualified type of MyFoo in order to know how much space is needed.

If you want to avoid typing (i.e. with fingers) the name of a particularly complex template, consider using a typedef:

typedef std::map<int, std::string> StringMap;

Or in C++0x you could use the auto keyword to have the compiler use type inference--though many will argue that leads to less readable and more error-prone code, myself among them. ;p

Answer 6

It makes a lot of sense it is like this, as Foo is not a class, only Foo<T> where T is a type.

In C++0x you can use auto, and you can create a function to make you a Foo, let's call it foo (lower case f). Then you would do

template<typename T> Foo<T> foo(int x)
{
  return Foo<T>(x);
}

auto myFoo = foo(55);