标签:方案 memory res bytes char har 解决方案 out 复制
前几天看到一个帖子,讨论两个记录复制,贴主提出了一个解决方案,但是不想用,希望大家集思广益,让人没想到的是方法还真的挺多,这里罗列一下。
这是需要复制的的两种记录
TypeA = record value1 : word; value2 : word; value3 : word; end; TypeB = record b1 : byte; b2 : byte; end;
方案1:贴主抛的砖1
CopyMemory(@TypeA, @TypeB, Length(TypeB))
方案2:贴主抛的砖2
TypeB.b1 := TypeA.value1 && $FF;
方案3.0:用变体记录
.. TypeAB = packed record case boolean of false:(a:TypeA); true:(b:TypeB); end; ..
方案3.1:用变体记录是低级形式,需要调整记录的定义形式,否则会由于系统不同上面的结构会产生不同的情况(确实存在BUG)。
type TTwoBytes : packed record ByteA, ByteB : Byte; end; TValueRecord : packed record: case Boolean of True: (Value: SmallInt); False: (ValueBytes : TTwoBytes); end; end; TMyRecord = packed record Value1, Value2, Value3 : TValueRecord; end; .... var MyRecord: TMyRecord; MyBytes: TTwoBytes; begin MyRecord := ...; // Fill it with data here // Access the words / smallints by something like: MyRecord.Value1.Value MyBytes := MyRecord.ValueBytes; // The key bit: type safe assignment // Do something with MyBytes.ByteA or MyBytes.ByteB end;
方案4:利用pascal自带的 Move
procedure CopyAtoB( const A: TypeA; var B: TypeB); begin // Assume A is bigger than B. Move( A, B, SizeOf( TypeB)) end; //or (with Math unit) procedure CopyAtoB( const A: TypeA; var B: TypeB); begin // No assumptions about A, B sizes. FillChar( B, SizeOf( B), 0); Move( A, B, Min( SizeOf( TypeA), SizeOf( TypeB))) end;
方案5:利用操作符重载
type TTypeA = record value1 : integer; value2 : integer; value3 : integer; end; TTypeB = record b1 : byte; b2 : byte; class operator Implicit(value : TTypeA):TTypeB; end; class operator TTypeB.Implicit(value: TTypeA): TTypeB; begin result.b1 := Hi(value.value1); result.b2 := Lo(value.value1); end; var a : TTypeA; b : TTypeB; begin b := a; end.
方案6:利用字节流复制
var a : TTypeA; bs : TBytesStream; bArr : TArray<byte>;//array of byte; begin bs := TBytesStream.Create(); bs.Write(a, sizeof(a)); bArr := bs.Bytes; end;
方案6:CopyMemory() (Windows unit)
type PTypeA = ^TTypeA; TTypeA = record value1 : word; value2 : word; value3 : word; end; PTypeB = ^TTypeB; TTypeB = record b1 : byte; b2 : byte; end; var A: TTypeA = (value1 : 11; value2 : 22; value3 : 33); B: TTypeB; B1: TTypeB; C: {packed?} array of Byte; begin Assert(SizeOf(TTypeA) >= SizeOf(TTypeB)); //... B:= PTypeB(@A)^; //... SetLength(C, SizeOf(A)); // TTypeA record to Byte Array PTypeA(@C[0])^:= A; // Byte Array to TTypeB B1:= PTypeB(@C[0])^ end;
标签:方案 memory res bytes char har 解决方案 out 复制
原文地址:https://www.cnblogs.com/hieroly/p/12334214.html