error: use of undeclared identifier 'T'

Question

I am new to using templates with structures. My main goal is to have auto variable inside structure.

More importantly, I am using a library function of Funct.Methd(const auto, &Handler). Here as you can see, in the place of the first argument, I would like to pass (based on some criteria) any const auto variable that is created by the user. I am expecting to have more than 20 variables, based on some criteria any 5 would be used.

After doing some research, I found that I can use template for this purpose. Please find the sample code below:

#include <iostream>
#include <string>

template<typename T>
struct FTmr {
    T query;
};

int main()
{
    FTmr<T> df;    
}

I am getting an error use of undeclared identifier 'T' when using the template with structures. First I need to resolve the above error and then I wanted to use const auto.

I was hoping to instantiate FTmr with df name and then access query variable.

df.query=<<const auto value>>

I would appreciate any help.

Answer 1

All variables in C++ have fixed types from the moment they are declared.

Templates are not types. Templates are instructions on how to make a type.

A const auto variable has a type determined by what you assign to it:

const auto x = foo();

the type of x is fixed, it is deduced by the fixed return type of foo() . The language just finds it for you.

A member of a struct is a variable. It must have its type determined within the struct the moment it is declared. You cannot defer its type until it is used.

struct Foo {
  static const auto x = 3;
};

that works because x 's type can be determined at the spot where x is declared.

In short, const auto doesn't work like that.

It may be possible to do what you really want to do, but it might be hard , and you probably lack the vocabulary to describe what it is you really want. My advice would be to not fight against the language and go with a fixed type.

Answer 2

What is T in main? You have to define it before using it. The previous T is scoped to the templated struct above. Also you cannot use that way; You have to instantiate the templated-struct this way:

FTmr<int> df;
FTmr<char> df;
// ...

T is the templated parameter which means as if passing a variable to a function it means passing the type instead. So the compile will create an instance of the struct with the type passed int.

• Functions take variables as parameters this way:

   void foo(int x, char y, double z, struct foo& the foo, long* pBar, ...); 

• Templates take types instead as parameters:

   template< class T> void Power(T& x); template < typename U, typename V> class Baz{ U _uval; V _vVal; }; 

So when instantiating the templated class / function the compiler will create an instance with that type:

    Baz<int, std::string> bistr;

An instance created by the compiler:

class Baz{
    int _uval;
   std:: string _vVal;
};

• Keep in mind that at compile time there is no template or type T but instead a specific version of the class is created.

Answer 3

tl;dr: Templates don't read your mind. You still have to tell them what to do.

const auto is not a "variable type". auto is a keyword that enables the deduction of a type from some information that you give it. For example, it knows that 5 is an int (because them's the rules), so auto x = 5; is possible.

In this case, you simply did not give it any information. How do you think it should know what to use for T ? We don't even know! Do you know? What should query be? What should it do? Think it through, then let your computer know what you've decided.