将32位指令拆分为字节并使用c将结果移动到另一个地址(split 32 bit instruction into bytes and move result to another address us

将32位指令拆分为字节并使用c将结果移动到另一个地址(split 32 bit instruction into bytes and move result to another address using c)

我有一个32位指令，我希望分成四个字节。 假设说明如下：

yyyyyzzzzzzxxxxxx?????????

该指令是一个由四个无符号整数组成的单词。 y表示操作码，并且 是为未使用的空间。 我正在研制一台大型机器。

我想要发生的是将值从z + w移到a 。

我从来没有在C工作过，但我曾试图这样做。

下面是我如何阅读该单词，以便我可以打印出每个字节：

unsigned int a, b, c, o; w = instruction << 24; z = instruction << 16; x = instruction << 8; y = instruction;

这里我打印无符号值，只是为了检查结果是什么。

printf("%u\n", w); printf("%u\n", z); printf("%u\n", x); printf("%u\n", y); printf("\n"); regs[x] = instruction + instruction << 8;

如果我在这之后打印出regs[x]的值，那么我可以看到我现在有一个值，但这是做这件事的正确方法吗？ 当我这样做时，我是否设置了register = z + w ？

编辑Mabye我应该得到这样的位？

y = (inst >> 24) & 077; x = (inst >> 16) & 0xffff; z = (inst >> 8) & 0xffff; w = (inst) & 0xffff;

然后这样做：

regs[y] = z + w;

i have a 32 bit instruction that i wish to split into four bytes. Let say the instruction looks like this:

yyyyyzzzzzzxxxxxx?????????

The instruction is a word that consists of four unsigned ints. y represents the operation code, and ??? are for the unused space. I am working on a big-endian machine.

What I would like to happen is to move the values from z + w to a.

I have never worked in C before but I have tried to do it like this.

Here is how I read the word, just so I ca print out each byte:

unsigned int a, b, c, o; w = instruction << 24; z = instruction << 16; x = instruction << 8; y = instruction;

Here I print unsigned values, just to check what the result are.

printf("%u\n", w); printf("%u\n", z); printf("%u\n", x); printf("%u\n", y); printf("\n"); regs[x] = instruction + instruction << 8;

if I print out the values of regs[x] after this, then I can see that I has a value now, but is this the correct way of doing it? When I do like this, do I set the register = z + w?

EDIT Mabye i should get the bits like this?

y = (inst >> 24) & 077; x = (inst >> 16) & 0xffff; z = (inst >> 8) & 0xffff; w = (inst) & 0xffff;

and then do like this:

regs[y] = z + w;

最满意答案

如果您只想使用位的位置和计数，则可以通过设置下一位和减少（1 << 10）-1来构建一个9位的位掩码。 所以你的价值观是

#define MASK(n) ((1<<(n+1))-1) unsigned int w = instruction & MASK(9); unsigned int x = (instruction >> 9) & MASK(6); unsigned int z = (instruction >> 15) & MASK(6); unsigned int y = (instruction >> 21) & MASK(5);

所有值都下移。 所以如果你想结合z和w你将不得不

unsigned int zw = z<<9 | w;

因为w包含9位，或者

unsigned int wz = w<<6 | z;

因为z包含6位。

If you like to use only bit positions and counts you can build a bit mask of i.e. 9 bits with setting the next bit and decrement (1<<10)-1. So your values are

#define MASK(n) ((1<<(n+1))-1) unsigned int w = instruction & MASK(9); unsigned int x = (instruction >> 9) & MASK(6); unsigned int z = (instruction >> 15) & MASK(6); unsigned int y = (instruction >> 21) & MASK(5);

all values are down shifted. So if you like to combine z and w you will have to

unsigned int zw = z<<9 | w;

because w contains 9 bits, or

unsigned int wz = w<<6 | z;

because z contains 6 bits.