为什么以下代码编译

Question

#include<iostream>
using namespace std;
class Foo {
void Bar( void ) const ;
};
int main()
{
  Foo f;
  cout<<sizeof(f)<<endl;
}

I ran this on g++,it did not give me any compilation error. Also, it executed giving o/p 1 which is correct. But I was expecting, error during linking. Is this compiler dependent?

Answer 1

I can only imagine that you expected to get an error as Foo::Bar is not defined. The One Definition Rule in the standard requires only that used elements are defined. In your particular case, nothing in your program uses Foo::Bar , so the program does not need that definition.

Answer 2

This will link because there are no outstanding references to Foo::Bar, and there for its definition is not required. Had you actually tried to make a call such as f.bar() it would have given you the error.

Answer 3

There is no linker error, because all dependencies are resolved.

As soon as you call the method Bar() and don't define it, you will get a linker error. Because then you reference Bar() and the linker cannot resolve it.