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digitalmars.D.learn - casting int to uint and vice versa - performance

reply Michal Minich <michal.minich gmail.com> writes:
What is performance overhead for casting value of type int to uint and in 
the opposite direction. What happen when I cast int to int or uint to uint 
(for the simplicity in generic code). I suppose there is no difference, 
and it is in fact no-op for processor - it depends only on 
interpretation. Only when making other operations compiler may generate 
different instruction when he considers some value of either of types. Am 
I right?

Thank you.
Dec 03 2010
parent reply "Steven Schveighoffer" <schveiguy yahoo.com> writes:
On Fri, 03 Dec 2010 16:34:51 -0500, Michal Minich  
<michal.minich gmail.com> wrote:


 What is performance overhead for casting value of type int to uint and in
 the opposite direction. What happen when I cast int to int or uint to  
 uint



Nothing, there is no performance, type is strictly a compile-time concept.


 (for the simplicity in generic code). I suppose there is no difference,
 and it is in fact no-op for processor - it depends only on
 interpretation. Only when making other operations compiler may generate
 different instruction when he considers some value of either of types. Am
 I right?



Not sure what you mean by that last part...

-Steve
Dec 03 2010
next sibling parent reply Michal Minich <michal.minich gmail.com> writes:
On Fri, 03 Dec 2010 16:44:42 -0500, Steven Schveighoffer wrote:


 On Fri, 03 Dec 2010 16:34:51 -0500, Michal Minich
 <michal.minich gmail.com> wrote:
 

 What is performance overhead for casting value of type int to uint and
 in the opposite direction. What happen when I cast int to int or uint
 to uint



 Nothing, there is no performance, type is strictly a compile-time
 concept.
 

 (for the simplicity in generic code). I suppose there is no difference,
 and it is in fact no-op for processor - it depends only on
 interpretation. Only when making other operations compiler may generate
 different instruction when he considers some value of either of types.
 Am I right?



 Not sure what you mean by that last part...
 
 -Steve



just that i + i generates different instructions than ui + ui where int 
i; uint ui;

Not really important, because I'm currently interested in cast only.

Thank you.
Dec 03 2010
parent Don <nospam nospam.com> writes:
spir wrote:

 On Fri, 03 Dec 2010 17:49:47 -0500
 "Steven Schveighoffer" <schveiguy yahoo.com> wrote:
 

 just that i + i generates different instructions than ui + ui where int
 i; uint ui;

 Not really important, because I'm currently interested in cast only.

 Thank you.  



 ui+ui.  As far as I know, adding unsigned and signed is the same  
 instruction, but unsigned add just ignores the carry bit.



 That used to be true (at least on some machines, a long time ago).
 
 Denis
 -- -- -- -- -- -- --
 vit esse estrany ☣
 
 spir.wikidot.com
 



It's only multiply, divide, and right shift where the instructions are 
different.
Dec 03 2010
prev sibling next sibling parent "Steven Schveighoffer" <schveiguy yahoo.com> writes:
On Fri, 03 Dec 2010 16:53:04 -0500, Michal Minich  
<michal.minich gmail.com> wrote:


 On Fri, 03 Dec 2010 16:44:42 -0500, Steven Schveighoffer wrote:


 On Fri, 03 Dec 2010 16:34:51 -0500, Michal Minich
 <michal.minich gmail.com> wrote:


 What is performance overhead for casting value of type int to uint and
 in the opposite direction. What happen when I cast int to int or uint
 to uint



 Nothing, there is no performance, type is strictly a compile-time
 concept.


 (for the simplicity in generic code). I suppose there is no difference,
 and it is in fact no-op for processor - it depends only on
 interpretation. Only when making other operations compiler may generate
 different instruction when he considers some value of either of types.
 Am I right?



 Not sure what you mean by that last part...

 -Steve



 just that i + i generates different instructions than ui + ui where int
 i; uint ui;

 Not really important, because I'm currently interested in cast only.

 Thank you.



That is odd, I would think that i+i generates the same instructions as  
ui+ui.  As far as I know, adding unsigned and signed is the same  
instruction, but unsigned add just ignores the carry bit.

-Steve
Dec 03 2010
prev sibling parent spir <denis.spir gmail.com> writes:
On Fri, 03 Dec 2010 17:49:47 -0500
"Steven Schveighoffer" <schveiguy yahoo.com> wrote:


 just that i + i generates different instructions than ui + ui where int
 i; uint ui;

 Not really important, because I'm currently interested in cast only.

 Thank you. =20



 That is odd, I would think that i+i generates the same instructions as =20
 ui+ui.  As far as I know, adding unsigned and signed is the same =20
 instruction, but unsigned add just ignores the carry bit.



That used to be true (at least on some machines, a long time ago).

Denis
-- -- -- -- -- -- --
vit esse estrany =E2=98=A3

spir.wikidot.com
Dec 03 2010