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?
Generally no, but in gcc You may make the last parameter of funcA() optional with a macro.
Default function parameters allow named parameters to be initialized with default values if no value or undefined is passed.
All the parameters of a function can be default parameters.
An IN OUT parameter cannot have a default value. An IN OUT actual parameter or argument must be a variable.
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); }
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