用Java读取VB 5.0二进制文件

发布于 2024-11-16 09:25:58 字数 884 浏览 3 评论 0原文

我有一个用 Visual Basic 5.0 编写的程序创建的二进制文件。该文件仅包含一堆来自 Visual Basic 世界的 Long 值。我知道 Visual Basic 5.0 中的 Long 大小为 4 个字节，但我不知道字节顺序。

我尝试使用各种“读取”方法通过 DataInputStream 解析文件，但我似乎得到“错误”（即负）值。

我怎样才能阅读并使用 Java 正确解释它？ Visual Basic 5.0 中 Long 的字节顺序是什么？

下面是我正在尝试使用的某种代码；我正在尝试读取 2 个 Long 并将其打印在屏幕上，然后再读取 2 个等。

    try {
        File dbFile = new File(dbFolder + fileINA);
        FileInputStream fINA = new FileInputStream(dbFile);
        dINA = new DataInputStream(fINA);
        long counter = 0;

        while (true) {
            Integer firstAddress = dINA.readInt();
            Integer lastAddress = dINA.readInt();

            System.out.println(counter++ + ": " + firstAddress + " " + lastAddress);
        }
    }
    catch(IOException e) {
        System.out.println ( "IO Exception =: " + e );
    }

原文

I have a binary file that is created from a program made in Visual Basic 5.0. The file just contains a bunch of Long values from the Visual Basic world. I've understood that Long in Visual Basic 5.0 is 4 bytes in size, but I do not know the byte order.

I tried parsing the file with DataInputStream using various "read"-methods, but I seem to get "wrong" (i.e. negative) values.

How can I read this and interpret it correctly with Java? What is the byte order for a Long in Visual Basic 5.0?

Below is some kind of code I'm trying to work with; I'm trying to read 2 Long s and print the out on the screen, and then read 2 more etc.

    try {
        File dbFile = new File(dbFolder + fileINA);
        FileInputStream fINA = new FileInputStream(dbFile);
        dINA = new DataInputStream(fINA);
        long counter = 0;

        while (true) {
            Integer firstAddress = dINA.readInt();
            Integer lastAddress = dINA.readInt();

            System.out.println(counter++ + ": " + firstAddress + " " + lastAddress);
        }
    }
    catch(IOException e) {
        System.out.println ( "IO Exception =: " + e );
    }

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半世晨晓 2024-11-23 09:25:58

由于 VB 在 x86 CPU 上运行，因此它的数据类型是小端字节序。另请注意，VB 中的 Long 与 Java 中的 int 大小相同。

我会尝试这样的事情：

int vbLong = ins.readUnsignedByte() +
             (ins.readUnsignedByte() << 8) +
             (ins.readUnsignedByte() << 16) +
             (ins.readUnsignedByte() << 24);

Since VB runs on x86 CPUs, its data types are little-endian. Also, note that a Long in VB is the same size as an int in Java.

I would try something like this:

int vbLong = ins.readUnsignedByte() +
             (ins.readUnsignedByte() << 8) +
             (ins.readUnsignedByte() << 16) +
             (ins.readUnsignedByte() << 24);

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