Optional non-const reference argument c++ ?

johnbach · #1 (**permalink**) 09-17-2009

Is it possible to have a non-const reference variable as an OPTIONAL/DEFAULT parameter to c++ function

Code:

ex
void read(string &data,int &type=0 /*or something*/) ;

so i will call

Code:

read(data);
//or
int type;
read(data,type);
printf("Type =%d",type);

I found one dirty workaround

Code:

#include<iostream>
#include<string>
using namespace std;
int dummy;
void read(string &data,int &type=dummy)
{
   data="TESTDATA";
   type=99;
}
int main(void)
{
   string val;

      int type;
      read(val,type);
      cout<<"TYPE:"<<type<<"\n";

      //or if i dont want the value of type
      read(val);
}

Any suggestion?

jim mcnamara · #2 (**permalink**) 09-17-2009

google for C++ variadic template - stdarg

You may also want to look at this -
C variadic function syntax:
Variadic Example - The GNU C Library

Gunther · #3 (**permalink**) 09-17-2009

Code:

void read(string &data,int &type=0 /*or something*/) ;

It might look cleaner and more concise, but remember that passing a variable by reference means to change its *actual* value. Since type, however, will be destroyed when read() ends, it wouldn't make too much sense to change its default value to something else. The change would be lost anyway.

Put a "const" in front of the "int" and it will compile. Why? Because it makes sense. A reference-to-const is read-only. There's nothing to be lost.

So, in conclusion. The "workaround" isn't really that dirty, because it keeps upright meaning of passing by reference.

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