为什么复制构造函数调用其他类的默认构造函数？

Question

I was wondering why an error like this would occur.

no matching function for call to 'Foo::Foo()'

in code for a copy constructor? Assume Foo is just an object with normal fields ( no dynamically allocated memory, etc. ), and the only constructor it has defined is a constructor that takes one argument.

I didn't even know the constructor needed to be considered though. If the code says something like

bar = thing.bar; //

and bar is of Foo type, with the specifications described above, shouldn't it just generate a shallow copy and be done with it? Why does a default constructor need to be invoked?

Answer 1

如果未定义构造函数，则编译器将生成默认构造函数，但是，如果您确实定义了构造函数（如副本构造函数），则编译器将不会生成默认构造函数，因此您也需要定义该构造函数。

Answer 2

It sounds like you've defined the copy constructor without defining any other constructor.

Once you declare an constructor explicitly, the compiler no longer provides a default constructor for you. Consequently, you no longer have a mechanism to construct an object of the class in the first place (and therefore wouldn't be able to copy it).

Answer 3

If, as you say, you're doing "something like

bar = thing.bar;

it's presumably in the body of your class's copy ctor -- so the bar field gets initialized with its class's default ctor first, then uses that class's assignment operator for this statement. If bar 's class only has a copy ctor, no default ctor, you'll need to add a bar(thing.bar) clause before your class's copy ctor opening { and remove that assignment (generally a good idea anyway, but mandatory under the "no default ctor" condition).