Question

假设指令aw是由32位结构定义的代码010，如下所示：

bits 31-25  unused (all 0s)
bits 24-22: code
bits 21-19: argument 1
bits 18-16: argument 2
bits 15-0:  offset (a 16-bit, 2's complement number with a range of -32768 to 32767)

鉴于号码8454151，如何确定代码是aw？

我试图将数字转换为22位，例如8454151＆gt;＆gt; 22，但我一直得到0.有关如何获取代码的位信息的任何想法（检查它是aw还是其他的东西）？

Answer 1

如果您只需验证指令是否属于某个操作，则需要最少循环的代码如下：

const uint32_t mask = 7 << 22;    // Shift 3 set bits by 22 in order to build 
                                  // a mask where bits 22-24 are set.
const uint32_t inst_aw = 2 << 22; // Shift the opcode by 22 to build a comparable value

uint32_t instruction = ...;       // Your instruction word

if ((instruction & mask) == inst_aw) {
  // Do your thing
}

无论如何你必须构建类似“指令解码器”或“解释器”的东西，我建议使用带有指令代码索引的指令名（或函数指针）的查找表：

/*! \brief Instruction names lookup table. Make sure it has always 8 entries 
           in order to avoid access beyond the array limits */
const char *instruction_names[] = {
  /* 000 */ "Unknown instruction 000",
  /* 001 */ "Unknown instruction 001",
  /* 010 */ "AW",
  ...
  /* 111 */ "Unknown instruction 111"
};


uint32_t instruction = ...;
unsigned int opcode = (instruction >> 22) & 0x07; // Retrieving the opcode by shifting right 22 
                                                  // times and mask out all bit except the last 3

printf("Instruction is: %s", instruction_names[opcode]);

Answer 2

有点转变应该有效。务必检查数据类型。您也可以使用0x01C000“和”数字然后进行比较。

从十进制获取有关二进制的信息

2 个答案: