char*(数组)转换为unsigned long + 1？

Question

我正在尝试将一些C代码移植到Java,但是我一直在努力弄清楚这些代码行的作用.

注意:rawEntry是类型的,char*并且看起来像是这样分配的12字节长

char *rawEntry = (char*)malloc(0x000c);

---

unsigned long *l;
unsigned long offset;

// ...

l = (unsigned long*) rawEntry + 1;
offset = ntohl(*l);

据我所知,它需要数组的前四项,并将它们组合在一起形成一个长的,但是我在java中的尝试并不成功.

offset = (rawEntry[0] << 24) +
         (rawEntry[1] << 16) +
         (rawEntry[2] << 8) +
         (rawEntry[3]) +
         1;

当出现以下数组时,

1 0 0 0 0 0 0 0 0 0 11 -38

C代码输出3034作为偏移量
My Java代码输出16777217,如果我翻转字节序则输出1

Answer 1

这个表达

l = (unsigned long*) rawEntry + 1;

强制转换rawEntry为指向系统上大小为8字节的类型的指针.之后,添加1意味着添加8个字节,因此实际转换如下所示:

offset = (Byte.toUnsignedInt(rawEntry[ 8]) << 24) +
         (Byte.toUnsignedInt(rawEntry[ 9]) << 16) +
         (Byte.toUnsignedInt(rawEntry[10]) <<  8) +
         (Byte.toUnsignedInt(rawEntry[11]) <<  0);

以下四个字节被解释为3034:

0 0 11 -38

您应该能够通过以下方式进一步简化ByteBuffer:

int offset = ByteBuffer.wrap(rawEntry, 8, 4).getInt();