The following code generates a compile error in Xcode:
template <typename T>
struct Foo
{
Foo(T Value)
{
}
};
int main()
{
Foo MyFoo(123);
return 0;
}
error: missing template arguments before 'MyFoo'
Changing Foo MyFoo(123);
to Foo<i开发者_开发问答nt> MyFoo(123);
fixes the issue, but shouldn't the compiler be able to figure out the appropriate datatype?
Is this a compiler bug, or am I misunderstanding implicit template parameters?
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
compiler can figure out template parameter type only for templated functions, not for classes/structs
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
In C++11 you can use decltype
:
int myint = 123;
Foo<decltype(myint)> MyFoo(myint);
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.
What you are trying to do now works in C++ 17. Template parameters can be inferred in C++ 17.
template <typename T>
struct Foo
{
Foo(T Value)
{
}
};
int main()
{
Foo a(123);
Foo b = 123;
Foo c {123};
return 0;
}
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);
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