Question

我正在尝试构建一个简单的程序集模拟器。我正在尝试创建set函数;该函数将采用2个参数（字符串数组）arg1＆amp; ARG2。然后它会将字符串与一个字符串数组进行比较，这是一个函数指针的arraylist的索引。

我遇到的问题是当我尝试设置寄存器的值时。我尝试过该系列的许多变种：

*register_ptr[i] = *(int*)(atoi(arg2));

没有成功;有什么我不理解的吗？

希望代码更清晰：

int opcode_set(char* opcode, char *arg1, char *arg2)
    {
      int i, j;
      //printf("%d\n", (int)arg2);
      for(i=0;i<4;i++)
      {
        if(!strcmp(arg1, register_str[i]))
        {
          for(j=0;j<4;j++)
          {
            if(!strcmp(arg2, register_str[i]))
            {
              *register_ptr[i] = *(int*)register_ptr[j];
            }
            else
            {
              *register_ptr[i] = *(int*)(atoi(arg2));

            }
          }
        }
      }
      //printf("%d", *register_ptr[i] );
      INSP++; /* NOP does not do anything except moving the instruction pointer to the next instruction */
      return (0); 
    }

编辑： register_str＆amp;的声明register_ptr：

 const char *register_str[] = {"REGA", "REGB", "REGC", "REGX"};
 int *register_ptr[]={&REGA, &REGB, &REGC, &REGX};

我使用两个数组来决定哪个操作码，一个字符串数组和一个函数指针数组，我通过字符串索引并使用相同的索引位置来调用该函数：

int exec_instruction(char *instruction){
int i; //used for indexing 

/* the line below may be useful for debugging to see the current instruction*/
/*printf("executing line: %s", instruction);*/

/* three variables could be used to extract opcode and 
 arguments from the variable instruction */
char *opcode = NULL;
char *arg1 = NULL;
char *arg2 = NULL ;
char *copy = NULL;

/* we need here some functionality to extract opcode, 
 arg1 and arg2 from the string instruction*/
copy = strdup(instruction);
opcode = strtok(copy," \n\r");
arg1 = strtok(NULL," \n\r");
arg2 = strtok(NULL," \n\r");


/* Now we have to call the right function corresponding 
 to the right opcode For example: */
for(i = 0; i < 10; i++)
{
  if(!strcmp(opcode_str[i], opcode))
    (*opcode_func[i])(opcode,arg1,arg2);
}

/* for demonstration purpose we execute NOP independent of input
 this line must go in your implementation */
(*opcode_func[0])("NOP",NULL,NULL);

/* return value should be 0 if execution was ok otherwise -1*/
return(0);
}

我说的两个数组：

    const char *opcode_str[] = {"NOP", "SET", "AND", "OR", "ADD", "SUB", "SHL", "SHR", "JMP", "PRT"};

    opcode_function opcode_func[] = { &opcode_nop, &opcode_set, &opcode_and, &opcode_or, &opcode_add, &opcode_sub, &opcode_shl, &opcode_shr, &opcode_jmp, &opcode_prt };

Answer 1

鉴于register_ptr的声明，行：

          *register_ptr[i] = *(int*)register_ptr[j];
        }
        else
        {
          *register_ptr[i] = *(int*)(atoi(arg2));

都是错的（一个是无害的，一个是无害的）。第一个不需要演员;你写得很好：

          *register_ptr[i] = *register_ptr[j];

第二个真的不需要任何转换，但它也不需要间接级别：

          *register_ptr[i] = atoi(arg2);

这会将atoi(arg2)返回的整数分配给register_ptr[i]所指向的内存，该内存可能是REGA，REGB，REGC或{ {1}}。如上所述，您将REGX中的值视为模拟器内存空间中的地址并读取其中的值，以及各种（可能）不需要的后果（例如核心转储）。

尝试将int传递给指针时出现总线错误

1 个答案: