• Benjamin Shropshire (15/15) Aug 14 2009 The same expression twice gets different results
• Jesse Phillips (2/25) Aug 14 2009 The value of 'b' is assigned during compile time, but since it is indire...
• BCS (4/29) Aug 14 2009 The only other option (beside saying this is correct) would be to make i...
• Jeremie Pelletier (3/36) Aug 14 2009 I don't think this is a good idea, since overload sets can be hijacked t...
• Jeremie Pelletier (2/29) Aug 14 2009 Exactly, when 'b' is declared, Fn(int) isn't declared yet, as it depends...

Benjamin Shropshire <ao pathlink.com> writes:

The same expression twice gets different results

code:

import std.stdio;

bool Fn(float i){ return true; }
const bool b = Fn(cast(int)0);

static if(b) bool Fn(int i){ return false; }
const bool c = Fn(cast(int)0);

void main()
{
   writef("%s\n", b);
   writef("%s\n", c);
}

output:

true
false

Jesse Phillips <jessekphillips+d gmail.com> writes:

Benjamin Shropshire Wrote:

 The same expression twice gets different results
 
 code:
 
 import std.stdio;
 
 bool Fn(float i){ return true; }
 const bool b = Fn(cast(int)0);
 
 static if(b) bool Fn(int i){ return false; }
 const bool c = Fn(cast(int)0);
 
 void main()
 {
    writef("%s\n", b);
    writef("%s\n", c);
 }
 
 output:
 
 true
 false
 

The value of 'b' is assigned during compile time, but since it is indirectly
called the behavior seems odd. I suppose it is something to be aware of, but it
is behaving correctly.

BCS <ao pathlink.com> writes:

Reply to Jesse,

 Benjamin Shropshire Wrote:
 
 The same expression twice gets different results
 
 code:
 
 import std.stdio;
 
 bool Fn(float i){ return true; }
 const bool b = Fn(cast(int)0);
 static if(b) bool Fn(int i){ return false; }
 const bool c = Fn(cast(int)0);
 void main()
 {
 writef("%s\n", b);
 writef("%s\n", c);
 }
 output:
 
 true
 false

 The value of 'b' is assigned during compile time, but since it is
 indirectly called the behavior seems odd. I suppose it is something to
 be aware of, but it is behaving correctly.
 

The only other option (beside saying this is correct) would be to make it 
illegal to add a new function to an overload set after the set is used. That 
could get very tricky to implement.

Jeremie Pelletier <jeremiep gmail.com> writes:

BCS Wrote:

 Reply to Jesse,
 
 Benjamin Shropshire Wrote:
 
 The same expression twice gets different results
 
 code:
 
 import std.stdio;
 
 bool Fn(float i){ return true; }
 const bool b = Fn(cast(int)0);
 static if(b) bool Fn(int i){ return false; }
 const bool c = Fn(cast(int)0);
 void main()
 {
 writef("%s\n", b);
 writef("%s\n", c);
 }
 output:
 
 true
 false

 The value of 'b' is assigned during compile time, but since it is
 indirectly called the behavior seems odd. I suppose it is something to
 be aware of, but it is behaving correctly.
 

 
 The only other option (beside saying this is correct) would be to make it 
 illegal to add a new function to an overload set after the set is used. That 
 could get very tricky to implement.
 
 

I don't think this is a good idea, since overload sets can be hijacked there is
nothing that prevents a module from using a local overload set, and another
then using it alongside its local hijacks.

Having such a "feature" would only limit the programmer in what they can do.

Jeremie Pelletier <jeremiep gmail.com> writes:

Jesse Phillips Wrote:

 Benjamin Shropshire Wrote:
 
 The same expression twice gets different results
 
 code:
 
 import std.stdio;
 
 bool Fn(float i){ return true; }
 const bool b = Fn(cast(int)0);
 
 static if(b) bool Fn(int i){ return false; }
 const bool c = Fn(cast(int)0);
 
 void main()
 {
    writef("%s\n", b);
    writef("%s\n", c);
 }
 
 output:
 
 true
 false
 

 
 The value of 'b' is assigned during compile time, but since it is indirectly
called the behavior seems odd. I suppose it is something to be aware of, but it
is behaving correctly.

Exactly, when 'b' is declared, Fn(int) isn't declared yet, as it depends on b
being true, int is implicitly convertible to float so it matches the first
declaration. When 'c' is declared both functions are available and the second
declaration is chosen.