Not possible: this pointer as a default argument. Why?

Question

The following code won't compile. Why?

class A
{
   int j;
   void f( int i = this->j );
}

Edit, for clarity. This is what I was trying to do, using less lines of code...

class A
{
   void f( int i ){};
   void f( );
   int j;
};

void A::f()
{
    f( j );
}

Because the instance is not in scope inside the argument declaration...? — Kerrek SB, Oct 10 '12 at 20:59
"...The expression can combine functions that are visible in the current scope, constant expressions, and global variables. The expression cannot contain local variables or non-static class-member variables..." — Adriano Repetti, Oct 10 '12 at 21:02
@Jordan Kaye - I have edited, to answer your question... I was trying to use less lines of code... as described above. — alexandreC, Oct 10 '12 at 21:07

score 11 · Accepted Answer · edited May 23 '17 at 12:29

11

Default argument values are bound at compile time.

"this" is only defined at run time, so can't be used.

See here for a fuller explanation: Must default function parameters be constant in C++?

edited May 23 '17 at 12:29

Community

1
1

answered Oct 10 '12 at 21:00

Eric

3,142
17
14

but why was C++ designed like this, that default arguments have to be bound at compiled time?... – alexandreC Oct 10 '12 at 21:11
@alexandreC that's another question. – Luchian Grigore Oct 10 '12 at 21:11

score 0 · Answer 2 · answered Oct 10 '12 at 21:10

Others have already commented on the reason this doesn't work. From one of the comments:

"...The expression can combine functions that are visible in the current scope, constant expressions, and global variables. The expression cannot contain local variables or non-static class-member variables..."

You could use optional to eliminate the extra function although I'm not sure it's clearer:

void f( boost::optional<int> i = boost::none ) { if(!i) i = j; ... }

Not possible: this pointer as a default argument. Why?

2 Answers2

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