Can you overload the sizeof operator in C++? Why or why not?

Can you overload the sizeof operator?

No, you cannot overload the sizeof operator in C++. This is because a compile-time operator that is handled directly by the compiler to determine the size, in bytes, of a data type or object. It is not treated as a function or a method that can be redefined.

Why cannot be overloaded:

Compile-Time Evaluation:
This is evaluated during compilation, not at runtime. Overloading operators in C++ applies to runtime behavior, which means the mechanism to overload would conflict with its compile-time purpose.

Fundamental Operator:
sizeof is a fundamental part of the language and is tightly integrated with the compiler. Allowing it to be overloaded could lead to inconsistent behavior or ambiguity in determining sizes.

Guaranteed Behavior:
The behavior is consistent and predictable. For instance:

int x;
std::cout << sizeof(x); // Always outputs the size of int on your platform

Overloading it could introduce uncertainty, defeating its primary purpose of providing reliable size information.

YouTube

Alternatives Exist:
If you need custom size-related behavior for specific types, you can define functions or templates:

template <typename T>
size_t getCustomSize(const T& obj) {
    return sizeof(obj) + customOffset;
}

Conclusion

This is intentionally designed to be non-overloadable to maintain its compile-time efficiency, simplicity, and reliability. If you need custom behavior, use functions or templates, but remember that these will work at runtime, not at compile time.