使用movememory复制数据 (2000年7月7日).txt

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使用MoveMemory复制数据 (2000年7月7日) 

本站更新  分类:   作者:srw  推荐:   阅读次数:885  
(http://www.codesky.net)  

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MoveMemory(lpvDest,lprSource,cbCopy)
MoveMemory参数意义:(1)lpvDest:目的数据的地址；(2)lpvSource:来源数据的
地址;(3)cbCopy:数据的字节数



var rect1,RECT2:TRect;
begin
rect1.Left:=10;
rect1.Top:=10;
rect1.Bottom:=200;
rect1.Right:=200;

rect2.Left:=50;
rect2.top:=100;
rect2.Bottom:=400;
rect2.Right:=390;
form1.Canvas.FillRect(rect1);
showmessage('movemory!');
form1.Refresh;

windows.MoveMemory(@rect1,@rect2,sizeof(TRect));
form1.Canvas.FillRect(rect1);

结果把rect2的值赋到rect1中


移动数组的值
一般的办法:

var s1,s2:array[0..4] of integer;
i:integer;
begin
s1[0]:=1;
s1[1]:=2;
s1[2]:=3;
s1[3]:=4;
s1[4]:=5;

s2[0]:=6;
s2[1]:=7;
s2[2]:=8;
s2[3]:=9;
s2[4]:=0;
for a:=0 to 4 do s1[a]:=s2[a];
end;

用movemeory的办法:

var s1,s2:array[0..4] of integer;
i:integer;
begin
s1[0]:=1;
s1[1]:=2;
s1[2]:=3;
s1[3]:=4;
s1[4]:=5;

s2[0]:=6;
s2[1]:=7;
s2[2]:=8;
s2[3]:=9;
s2[4]:=0;
for i:=0 to 4 do showmessage('s1的'+inttostr(i)+':'+inttostr(s1[i]));
showmessage('movemeory1!');
windows.MoveMemory(@s1,@s2,sizeof(s1));

for i:=0 to 4 do showmessage('s1的'+inttostr(i)+':'+inttostr(s1[i]));