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当我结合两个 32 位整数时，它应该形成一个 64 位长整数，但它仍然是一个 32 位负值。怎么了？

仅这一行就应该生成比 32 位更大的数字：(readDword() And &HFFFFFFFF) << 32)

用于测试的 VB.NET 代码

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click

    Dim u As Long = ((readDword() And &HFFFFFFFF) << 32) Or _
        (readDword2() And &HFFFFFFFF)
End Sub

Function readDword() As Integer
    Dim i As Integer = (((0 And &HFF) << 24) Or _
((0 And &HFF) << 16) Or _
((11 And &HFF) << 8) Or _
(58 And &HFF))
    Return i
End Function

Function readDword2() As Integer
    Dim i As Integer = (((241 And &HFF) << 24) Or _
((145 And &HFF) << 16) Or _
((136 And &HFF) << 8) Or _
(247 And &HFF))
    Return i
End Function



预期结果
-----------------
输入字节 =0,0,11,58,241,145,136,247
我得到的结果 =-242119681
预期结果 = 12347788855543
--------- ---------------------


这可以正常工作，但它是倒退的，所以我尝试自己按位进行

    Dim bytesz() As Byte = New Byte() {247, 136, 145, 241, 58, 11, 0, 0}
    Dim a As Int64 = BitConverter.ToInt64(bytesz, 0)

字节是如何用 Java 编写的

  public void writeQWord(long l) throws IOException {
    write((int) (l >> 56)); //0
    write((int) (l >> 48)); //0
    write((int) (l >> 40)); //11
    write((int) (l >> 32)); //58
    write((int) (l >> 24)); //241
    write((int) (l >> 16)); //145
    write((int) (l >> 8));  //136
    write((int) (l));   }   //247

固定的

正确的代码是：

Public Function ReadQWord() As Long
    Return (CLng(ReadDWord() And &HFFFFFFFFL) << 32) Or CLng(ReadDWord() And &HFFFFFFFFL)
End Function
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1 回答 1

1

使用按位运算符和常量时，请确保常量类型反映您的使用情况。

这个

    Dim u As Long = 1 << 32

和这个

    Dim u As Long = 1L << 32

产生不同的结果。我还注意到你的函数 readDword 只产生一个整数作为返回和内部。

于 2013-07-06T13:36:24.807 回答