简介

本系列为 $2023$ 级计科中英 《计算机组织与架构》 代码自测题。如果你不是 $2023$ 级计科中英的同学，请忽略本题。

题目已经包含大作业所给出的测试输入输出，你只需要把题目代码中的 fopen("./test1.txt", "r") 改为 stdin 即可正常在此测试代码。

本题题目所给代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//This is an array of register mnemonics in y86
const char *register_names[] =
    {
        "%eax",
        "%ecx",
        "%edx",
        "%ebx",
        "%esp",
        "%ebp",
        "%esi",
        "%edi",
        "UNKNOWN_REGSITER"};

int convertStrToByteCode(const char *str, unsigned char inst[], int size);

int main(int argc, char **argv)
{
  FILE *pFile = NULL;

  char buffer[15];

  if (argc < 2)
  {
    pFile = stdin;
  }
  else
  {
    pFile = fopen(argv[1], "r");
  }

  if (pFile == NULL)
  {
    printf("Error open test file, please make sure they exist.\n");

    return 0;
  }

  while (fgets(buffer, 15, pFile) && strlen(buffer) > 1)
  {
    //This unsigned char array stores an instruction read from the file
    //As the largest y86 instruction is 6 bytes, there are 6 unsigned char in the array where
    //each represents a byte.
    unsigned char instruction[6] = {0, 0, 0, 0, 0, 0};
    convertStrToByteCode(buffer, instruction, 6);

    //TODO: From here, your task is to complete the implementation so that all y86 opcode and operands can be disassembled.
    //Any undisassembled opcode should display as "TODO: undisassembled opcode and operands"
    printf("TODO: undisassembled opcode and operands. The first byte of the instruction is 0x%X\n", instruction[0]);
  }

  fclose(pFile);

  return 0;
}

/****************************************************************************
N.B. You do not need to modify or work in this function.
Description:
This function converts a line of machine code read from the text file
into machine byte code.
The machine code is stored in an unsigned char array.
******************************************************************************/
int convertStrToByteCode(const char *str, unsigned char inst[], int size)
{
  int numHexDigits = 0;
  char *endstr;
  //Each instruction should consist of at most 12 hex digits
  numHexDigits = strlen(str) - 1;
  //Convert the string to integer, N.B. this integer is in decimal
  long long value = strtol(str, &endstr, 16);

  int numBytes = numHexDigits >> 1;
  int byteCount = numHexDigits >> 1;

  while (byteCount > 0)
  {
    unsigned long long mask = 0xFF;
    unsigned long shift = (numBytes - byteCount) << 3;

    inst[byteCount - 1] = (value & (mask << shift)) >> shift;
    byteCount--;
  }

  //Return the size of the instruction in bytes
  return numHexDigits >> 1;
}