Why C++ does not allow function parameters used fo

问题:

This is a follow-up on this question. The code in the OP question there looked quite reasonable and unambiguous to me. Why does not C++ allow using former parameters to define default values of latter parameters, something like this:

int foo( int a, int b = a );

Also, at least in C++11 declared types of parameters can be used to determine the return type, so it's not unheard of to use function parameters in similar manner:

auto bar( int a ) -> decltype( a );

Thus the question: what are the reason(s) why the above declaration of foo is not allowed?

回答1:

For one thing, this would require that a is evaluated before b, but C++ (like C) does not define the order of evaluation for function parameters.

You can still get the effect you want by adding an overload:

int foo(int a, int b)
{ /* do something */ }

int foo(int a)
{ return foo(a, a); }