Here is the code
It's interesting to see the size of A1 and A2 are different just because you give A2.B a name! Why is THAT?Code:struct A1{ char b; struct { char c; char a; }; }; cout<<sizeof(A1)<<endl; struct A2{ char b; struct B{ char c; char a; }; }; cout<<sizeof(A2)<<endl;
You didn't actually declare struct B as a member of A2, so it's members are inaccessible and are probably eliminated by an optimizer. This would produce a smaller struct.
On the other hand, the anonymous struct in A1 is meaningless, so it has no impact on the accessability of its members. The optimizer probably scraps the anonymous wrapper.
The output is
I don't understand why a byte alignment is applied to A1?Code:3 1
Originally Posted by citizen
It's not an alignment issue. struct B is not a member of A2. Therefore, its members are not a part of the object, producing something of a smaller size.
If you run that, foo and quz might be the same size, but this is only because quz actually has a bar.Code:#include <iostream> struct foo { char stub[3]; }; struct bar { char stub[2]; }; struct quz { char c; bar b; }; int main ( ) { std::cout<<"sizeof (quz) = "<<sizeof (quz)<<"\n"; std::cout<<"sizeof (foo) = "<<sizeof (foo)<<"\n"; }
Last edited by whiteflags; 05-18-2008 at 09:18 PM.
Simply put, you've hit a compiler extension, so you shouldn't be surprised that it does weird things.Code:~/formus $ g++ -o anonstruct -ansi -pedantic -Wall anonstruct.cpp anonstruct.cpp:9: error: ISO C++ prohibits anonymous structs
All the buzzt!
CornedBee
"There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
- Flon's Law
