有没有一种优雅的方法可以用 Delphi (6) 删除文件中的奇偶校验位?在这种情况下,奇偶校验位是每第 9 位。
Ernest
问问题
780 次
2 回答
4
假设您的文件是一个包含 9 位块的长位流,并且您希望输出相同的流但具有 8 位块(即每 9 位丢弃一次)。
您可以一次读取 9 个字节(72 位 = 8 个 9 位块),然后使用位移将它们放入 8 个 8 位块中。
您需要一些特殊处理来处理不是 9 字节的倍数的文件,所以这只是一个粗略的指南。
procedure TForm1.Button1Click(Sender: TObject);
var
FSIn: TFileStream;
FSOut: TFileStream;
InBuffer: array[0..8] of Byte;
OutBuffer: array[0..7] of Byte;
X: Integer;
BytesRead: Integer;
BytesToWrite: Integer;
begin
FSIn := TFileStream.Create('Input.dat', fmOpenRead);
FSOut := TFileStream.Create('Output.dat', fmCreate);
try
for X := 1 to FSIn.Size div 9 do
begin
FillChar(InBuffer[0], 9, 0);
BytesRead := FSIn.Read(InBuffer[0], 9);
OutBuffer[0] := InBuffer[0];
OutBuffer[1] := (InBuffer[1] and 127) shl 1 + (InBuffer[2] and 128) shr 7;
OutBuffer[2] := (InBuffer[2] and 63) shl 2 + (InBuffer[3] and 192) shr 6;
OutBuffer[3] := (InBuffer[3] and 31) shl 3 + (InBuffer[4] and 224) shr 5;
OutBuffer[4] := (InBuffer[4] and 15) shl 4 + (InBuffer[5] and 240) shr 4;
OutBuffer[5] := (InBuffer[5] and 7) shl 5 + (InBuffer[6] and 248) shr 3;
OutBuffer[6] := (InBuffer[6] and 3) shl 6 + (InBuffer[7] and 252) shr 2;
OutBuffer[7] := (InBuffer[7] and 1) shl 7 + (InBuffer[8] and 254) shr 1;
if BytesRead < 9 then
begin
// To do - handle case where 9 bytes could not be read from input
BytesToWrite := 8;
end else
BytesToWrite := 8;
FSOut.Write(OutBuffer[0], BytesToWrite);
end;
finally
FSIn.Free;
FSOut.Free;
end;
end;
于 2009-06-09T16:15:05.310 回答
0
假设您可以将事物一对一地读入整数。
我 := 我 xor 512;
于 2009-06-10T08:12:15.477 回答