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* patch by Rika: improve ctuils.newalign/align, part of #39496

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florian 2021-12-23 22:00:26 +01:00
parent b4c8c1da12
commit ab969e0a9b
1 changed files with 29 additions and 34 deletions
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63
compiler/cutils.pas
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@ -293,13 +293,13 @@ implementation
function newalignment(oldalignment: longint; offset: int64): longint;
var
localoffset: longint;
begin
localoffset:=longint(offset);
while (localoffset mod oldalignment)<>0 do
oldalignment:=oldalignment div 2;
newalignment:=oldalignment;
{ oldalignment must be power of two.
Works even for negative offsets (but not alignments),
as two's complement of, say, 2^4+2^3 is 11000 and -2^4-2^3 is 11...1101000 -
both end with exactly N zeros, where N is the largest power of two that divides the number
(smallest power of two involved in these sums). }
result:=1 shl BsfQWord(qword(offset or oldalignment));
end;
@ -340,50 +340,45 @@ implementation
function align(i,a:longint):longint;{$ifdef USEINLINE}inline;{$endif}
{
return value <i> aligned <a> boundary
return value <i> aligned <a> boundary. <a> must be power of two.
}
begin
{ for 0 and 1 no aligning is needed }
if a<=1 then
result:=i
else
begin
if i<0 then
result:=((i+1-a) div a) * a
else
result:=((i-1+a) div a) * a;
end;
{ One-line formula for i >= 0 is
>>> (i + a - 1) and not (a - 1),
and for i < 0 is
>>> i and not (a - 1). }
if a>0 then
a:=a-1; { 'a' is decremented beforehand, this also allows a=0 as a synonym for a=1. }
if i>=0 then
i:=i+a;
result:=i and not a;
end;
function align(i,a:int64):int64;{$ifdef USEINLINE}inline;{$endif}
{
return value <i> aligned <a> boundary
return value <i> aligned <a> boundary. <a> must be power of two.
}
begin
{ for 0 and 1 no aligning is needed }
if a<=1 then
result:=i
else
begin
if i<0 then
result:=((i+1-a) div a) * a
else
result:=((i-1+a) div a) * a;
end;
{ Copy of 'longint' version. }
if a>0 then
a:=a-1;
if i>=0 then
i:=i+a;
result:=i and not a;
end;
function align(i,a:qword):qword;{$ifdef USEINLINE}inline;{$endif}
{
return value <i> aligned <a> boundary
return value <i> aligned <a> boundary. <a> must be power of two.
}
begin
{ for 0 and 1 no aligning is needed }
if (a<=1) or (i=0) then
result:=i
else
result:=((i-1+a) div a) * a;
{ No i < 0 case here. }
if a>0 then
a:=a-1;
result:=(i+a) and not a;
end;