CODEWITHSUNDEEP
c++c++11c++-chrono
Irinel Iovan
Irinel Iovan

Reputation: 895

Missing type specifier and one more question

I'm trying to make short chrono declarations

#pragma once
#include <chrono>

    class Foo//: public SINGLETON<Foo>
    {
    public:
        using point = std::chrono::steady_clock::time_point;
        using secs  = std::chrono::duration_cast<std::chrono::seconds>;
        using now   = std::chrono::steady_clock::now();
    };

So that's the whole code. First issue, this code don't want to compile. First compile error:

 error: 'duration_cast<std::chrono::seconds>' in namespace 'std::chrono' does not name a type
  using time_secs  = std::chrono::duration_cast<std::chrono::seconds>;

Second error;

error: expected type-specifier
  using time_now   = std::chrono::steady_clock::now();

Now a question about usage of those, i want to use them like this;

time_point  GetElapsedtTime(){return time_secs(time_now - Another_time_point ); }

Can I use them like above?

I'm trying to set time point intro a variable, then to countdown the elapsed time until it reaches 1hour for example, then when 1 hour elapses.. do something.

Upvotes: 0

Views: 114

Answers (1)

Marshall Clow
Marshall Clow

Reputation: 16670

using (in this context) is the equivalent of typedef. You are giving a type a new name.

This one: using point = std::chrono::steady_clock::time_point; is fine. std::chrono::steady_clock::time_point is a type; you're making a new name.

This one: using secs = std::chrono::duration_cast<std::chrono::seconds>; is not, because std::chrono::duration_cast<std::chrono::seconds> is not a type - it is a function.

Similarly, std::chrono::steady_clock::now(); is not a type. It is a value of type std::chrono::time_point<std::chrono::steady_clock>.

(which is what the "does not name a type" message that the compiler gave you says)

Upvotes: 1

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