Understanding Why a const Reference to Base Cannot Be Initialized with constexpr Derived

Explore the reasons behind the inability to initialize a `const reference to base` using a `constexpr derived` object in C+ + . We'll break down the concept and provide clarity on the importance of static storage.
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Why Can't a const Reference to Base Be Initialized from a constexpr Derived Object?

In C+ + , we often come across situations where we want to leverage the strengths of constexpr for improved performance and safety during compile-time. However, despite these advantages, developers can sometimes run into perplexing errors. One such scenario is trying to initialize a const reference to base from a constexpr derived object, which raises some eyebrows. This guide aims to unravel this complex problem by explaining what's happening under the hood and how to resolve it.

The Scenario: Initialization Error Explained

Consider the following example:

[[See Video to Reveal this Text or Code Snippet]]

In this code, we define two structs: a base struct and a derived struct that inherits from it. We then try to initialize a const base reference b with a constexpr derived object d. However, both G+ + and Clang provide errors indicating that the initialization does not yield a constant expression.

Error Messages from the Compilers

G+ + Error:
error: '(const base&)(& d)' is not a constant expression

Clang Error:
error: constexpr variable 'b' must be initialized by a constant expression.

note: reference to subobject of 'd' is not a constant expression. Add 'static' to give it a constant address.

From these errors, it's clear that the references to subobjects of d are not constant expressions, despite d having a constexpr constructor.

Understanding the Root Cause

The key here lies in understanding what constexpr means in C+ + . It guarantees that an expression can be evaluated at compile time, but it does not affect the storage duration of the variable.

1. Storage Duration of constexpr Variables

When you declare a non-static constexpr variable, it is stored in dynamic (stack) memory, just like any other non-static local variable. For instance:

[[See Video to Reveal this Text or Code Snippet]]

This means:

The address of a can change with each invocation of test(), as multiple instances can exist during recursion or when accessed by different threads.

Thus, a's address cannot be a constant expression since it is dynamically allocated on the stack.

2. Impact of static Declaration

When you declare a variable as static, it changes the storage allocation from stack to static memory. In simple terms, it promotes the variable to behave like a global variable, only within the local scope of the function. Here’s what happens when we use static:

The variable maintains a fixed address, assuring that there is only one copy of it, regardless of how many times the function is executed.

The address of a static variable is indeed a constant expression, which resolves the error you encountered.

Solution: Using static for Persistent Storage

To fix the initialization error you received in the original example, simply add the static keyword in front of the declaration of d in the following way:

[[See Video to Reveal this Text or Code Snippet]]

This adjustment ensures that d is allocated in static memory, allowing you to correctly initialize b because the reference's address is constant.

Conclusion

Understanding the intricacies of constexpr, storage duration, and the role of static storage can significantly aid in debugging common C+ + problems. By recognizing that a const reference to base must point to an object with a constant address, we can resolve the errors that may arise from trying to assign a constexpr derived object.

Remember, it's the details of memory management that often create barriers to what seems conceptually simple. Kee