我如何从C中32位序列的立即数中获取值? [英] How I get the value from the Immediate part of a 32 Bit sequence in C?

问题描述

我用C语言构建了一个虚拟机.为此，我有了指令

I built a virtual machine in C. And for this I have the Instruction

pushc <const>

我将命令和值保存在32位中.前8位用于命令，其余部分用于值.

I saved the command and the value in 32 Bit. The First 8 Bit are for the command and the rest for the value.

8位->操作码 24位->立即值

为此，我制作了一个宏

#define PUSHC 1 //1 is for the command value in the Opcode
#define IMMEDIATE(x) ((x) & 0x00FFFFFF)

更新:

**#define SIGN_EXTEND(i) ((i) & 0x00800000 ? (i) | 0xFF000000 : (i))** 

然后我加载以在无符号int数组中对此进行测试:

Then I load for testing this in a unsigned int array:

更新:

unsigned int code[] = { (PUSHC << 24 | IMMEDIATE(2)),
                        (PUSHC << 24 | SIGN_EXTEND(-2)),
                         ...};

稍后在我的代码中，我想获取pushc命令的即时值并将该值推入堆栈...

later in my code I want to get the Immediate value of the pushc command and push this value to a stack...

我从数组中获取每个指令(IR)并构建了我的堆栈.

I get every Instruction (IR) from the array and built my stack.

更新:

 void exec(unsigned int IR){

      unsigned int opcode = (IR >> 24) & 0xff;
      unsigned int imm = (IR & 0xffffff);

     switch(opcode){
       case PUSHC: {
         stack[sp] = imm;
         sp = sp + 1;
         break;
}
}

    ...

    }
    }

推荐答案

只需使用按位AND屏蔽掉低24位，然后在case中使用它即可:

Just use a bitwise AND to mask out the lower 24 bits, then use it in the case:

const uint8_t opcode = (IR >> 24) & 0xff;
const uint32_t imm = (IR & 0xffffff);
switch(opcode)
{
  case PUSHC:
    stack[sp] = imm;
    break;
}

我转移了操作码的提取内容，以使case易于阅读.

I shifted around the extraction of the opcode to make the case easier to read.

这篇关于我如何从C中32位序列的立即数中获取值?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！