C++结构体内存布局差异问询：含成员函数的结构体是否改变布局？

Great question! Let's break this down clearly for you.

First, the key point to remember: member functions (like B::foo()) do NOT take up space in individual instances of your struct.

Here's why: Member functions are part of the struct's "type information" stored in the program's code segment, not duplicated for every instance you create. When you call foo() on a B object, the compiler implicitly passes a pointer to that object (the this pointer) to the function—but that pointer isn't stored inside the object itself.

So, for your specific examples:

• The memory layout of a struct A instance is exactly the same as a struct B instance. Both will have the int a, double b, and float c members laid out in the same order, with the same padding (if any) for alignment, and the same total size.
• If you check sizeof(A) and sizeof(B) with a standard-compliant compiler, they'll return the exact same value.

Wait, what if we added a virtual function? Then things would change—struct instances would include a hidden pointer to a virtual function table (vtable), which adds to the size and alters the layout. But in your case, B::foo() is a non-virtual member function, so no extra overhead is added to the struct's memory footprint.

To sum up: Adding non-virtual member functions to a struct doesn't change how its data members are arranged in memory. Only data members (and virtual function pointers, if any) affect the instance's memory layout.

内容的提问来源于stack exchange，提问作者Iliketoproveit