Generalized Plain Old Data


Plain Old Data (POD) obeys the C standard layout. Therefore, you can directly apply the fast C functions memcopy, memmove, memset, or memcmp.



PODs are in classical C++ fundamental types like booleans, integers of floating-point numbers. The restriction will not hold for C++11. With C++11, even classes and structs can be PODs. For simplicity reasons, I only speak about classes. 

Which requirements hold for the C++11 class to be a POD? A class is a POD, if it's trivial, has a standard layout, and all of its non-static members are PODs. The definition is quite concise. But what does it mean that class should be trivial and has a standard layout?

Now the standard reads like German legal text.

Trivial class

A class is trivial if it  

  • has a trivial default constructor.
  • is trivially copyable.


A trivially copyable class is a class that

  • has no non-trivial copy or move constructor.
  • has no non-trivial copy or move assignment operator.
  • has a trivial destructor.

Non-trivial means that the developer implements the mentioned methods. The method is trivial if a method is requested from the compiler via the keyword default or automatically generated from the compiler.

The definition of a POD goes on with the standard layout.

Standard layout

A class has a standard layout if it has no

  • virtual functions.
  • virtual base classes.
  • references.
  • different access specifiers (public, protected, and private).

It's a lot easier to check with the help of the type-traits library if the class is POD.


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Checking types with the type-traits library

// pod.cpp

#include <iostream>
#include <type_traits>

struct Pod{
  int a;

struct NotPod{
    int i;
    int j;

int main(){

  std::cout << std::boolalpha << std::endl;
  std::cout << "std::is_pod<Pod>::value: " << std::is_pod<Pod>::value << std::endl;
  std::cout << "std::is_pod<NotPod>::value: " << std::is_pod<NotPod>::value << std::endl;
  std::cout << std::endl;
  std::cout << "std::is_trivial<NotPod>::value: " << std::is_trivial<NotPod>::value << std::endl;
  std::cout << "std::is_standard_layout<NotPod>::value: " << std::is_standard_layout<NotPod>::value << std::endl;
  std::cout << std::endl;


The class Pod in lines 6 - 8 is a POD, but not the class NotPod (lines 10 -15). We get the answer relatively easy with the help of the function std::is_pod (lines 21 - 22) from the type-traits library. But we can do even better with the type-traits library. I analyze in line 26 and 27 in the class NotPod even more. The result is: NotPod is trivial but has no standard layout. NotPod has no standard layout because the variable i is public. On the contrary, the variable j is private.

The output of the program depicts the explanation.


What's next?

This post finishes the series of posts about the features in C++ that are very important from the performance perspective. In the next post, I will continue my blog with posts about carefully handling resources. Memory management has a high priority in embedded development. Therefore, it fits very well that C++11 has the new smart pointers std::shared_ptr, std::unique_ptr, and std::weak_ptr, and the manual memory management with new becomes almost unnecessary.





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Tags: POD


0 #1 Marco 2016-12-04 18:04
Hi, just an observation. Technically, the only requirement for an object to be compatible with C layout is to be standard layout.

However, both trivially copyable and standard layout objects can be memcpy'd (and others).

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