• SKS (19/19) Oct 12 2006 I have a program:
• Derek Parnell (19/41) Oct 12 2006 No it is not a bug. The 'out' qualifier causes the argument to be set to
• Craig Black (6/10) Oct 12 2006 And I think that is gay. 'inout' should not be required to pass an argu...
• BCS (3/6) Oct 12 2006 IIRC that would be: only *out* and inout arguments get fed back to the
• Derek Parnell (6/14) Oct 12 2006 Of course ... just testing to see if you were awake ;-)
• he_the_great (2/10) Oct 16 2006 Good, I'm not crazy.

SKS <skshukla_ncr yahoo.co.in> writes:

I have a program:
import std.file;

public static void foo(in int i, out int o, inout int io)
{
	i++; o--; io++;
	printf("foo %d, %d, %d\n", i, o, io);
}

int main (char[][] args)
{
	int i1 = 1, i2 = 123, i3 = 234;
	printf("before foo %d, %d, %d\n", i1, i2, i3);
	foo(i1, i2, i3);
	printf("after foo %d, %d, %d\n", i1, i2, i3);
	return 0;
}

Why this compiles successfully with dmd compiler?
I expect 'out' parameter modification to cause error. Is it a bug in compiler
implementation or wrong usage?

Also looking at sample programs I don't find in/out usage as common?

Derek Parnell <derek psyc.ward> writes:

On Thu, 12 Oct 2006 08:46:32 +0000 (UTC), SKS wrote:

 I have a program:
 import std.file;
 
 public static void foo(in int i, out int o, inout int io)
 {
 	i++; o--; io++;
 	printf("foo %d, %d, %d\n", i, o, io);
 }
 
 int main (char[][] args)
 {
 	int i1 = 1, i2 = 123, i3 = 234;
 	printf("before foo %d, %d, %d\n", i1, i2, i3);
 	foo(i1, i2, i3);
 	printf("after foo %d, %d, %d\n", i1, i2, i3);
 	return 0;
 }
 
 Why this compiles successfully with dmd compiler?
 I expect 'out' parameter modification to cause error. Is it a bug in compiler
 implementation or wrong usage?

No it is not a bug. The 'out' qualifier causes the argument to be set to
the datatype's .init value before the function gets control. The 'inout'
qualifier allows the value of the argument to passed to the function by the
caller, just like 'in' does. Once the function gets control, it can modify
the passed arguments as much as it likes, but only changes to 'in' and
'inout' arguments get fed back to the calling code.
 
 Also looking at sample programs I don't find in/out usage as common?

That's because it is better to limit cohesion between caller and called
code. The use of 'in' and 'out' reduces the possibility of side-effects
that are unknown to the caller and reduces the need for the called function
to be responsible for changes to data it doesn't own.

When 'inout' is used, it is usually as a compromise to improve performance.
For example, it can be a performance cost to copy large structs or
fixed-length arrays betwen functions, and 'inout' is a way to avoid that.

-- 
Derek Parnell
Melbourne, Australia
"Down with mediocrity!"

"Craig Black" <cblack ara.com> writes:

 When 'inout' is used, it is usually as a compromise to improve 
 performance.
 For example, it can be a performance cost to copy large structs or
 fixed-length arrays betwen functions, and 'inout' is a way to avoid that.

And I think that is gay.  'inout' should not be required to pass an argument 
by reference.
IMO 'in' should pass by reference, but denote a read only parameter.  This 
would be better
for performance and correctness.

-Craig 

BCS <BCS pathlink.com> writes:

Derek Parnell wrote:
 Once the function gets control, it can modify
 the passed arguments as much as it likes, but only changes to 'in' and
 'inout' arguments get fed back to the calling code.

IIRC that would be: only *out* and inout arguments get fed back to the 
calling code.

Derek Parnell <derek psyc.ward> writes:

On Thu, 12 Oct 2006 09:05:23 -0700, BCS wrote:

 Derek Parnell wrote:
 Once the function gets control, it can modify
 the passed arguments as much as it likes, but only changes to 'in' and
 'inout' arguments get fed back to the calling code.

 
 IIRC that would be: only *out* and inout arguments get fed back to the 
 calling code.

Of course ... just testing to see if you were awake ;-)

-- 
Derek Parnell
Melbourne, Australia
"Down with mediocrity!"

he_the_great <jessekphillips+digitalmars gmail.com> writes:

BCS wrote:
 Derek Parnell wrote:
 Once the function gets control, it can modify
 the passed arguments as much as it likes, but only changes to 'in' and
 'inout' arguments get fed back to the calling code.

 
 IIRC that would be: only *out* and inout arguments get fed back to the 
 calling code.

Good, I'm not crazy.