C代码 - 需要澄清有效性

时间:2009-10-09 05:52:09

标签: c++ c unix gdb dbx

您好我已经根据要求编写了代码。

(field1_6)(field2_30)(field3_16)(field4_16)(field5_1)(field6_6)(field7_2)(field8_1 ) ..... 这是一个桶(8个字段)的数据。我们将一次收到20个桶,意味着总共160个字段。 我需要取field3,field7&的值。 fields8基于预定义的条件。 如果输入参数为N,则从第一个桶中取出三个字段,如果是我需要的Y. 从第一个以外的任何其他桶获取三个字段。 如果argumnet是Y,那么我需要一个接一个地扫描所有20个桶并检查 存储桶的第一个字段不等于0,如果为真,则获取该存储桶的三个字段并退出。 我写了代码,它也工作得很好..但不是那么自信,它是有效的。 我害怕崩溃了一段时间。请在下面提示代码。

int CMI9_auxc_parse_balance_info(char *i_balance_info,char  *i_use_balance_ind,char *o_balance,char *o_balance_change,char *o_balance_sign
)
{
  char *pch = NULL;
  char *balance_id[MAX_BUCKETS] = {NULL};
  char balance_info[BALANCE_INFO_FIELD_MAX_LENTH] = {0};
  char *str[160] = {NULL};
  int i=0,j=0,b_id=0,b_ind=0,bc_ind=0,bs_ind=0,rc;
  int total_bukets ;
  memset(balance_info,' ',BALANCE_INFO_FIELD_MAX_LENTH);
  memcpy(balance_info,i_balance_info,BALANCE_INFO_FIELD_MAX_LENTH);
  //balance_info[BALANCE_INFO_FIELD_MAX_LENTH]='\0';
  pch = strtok (balance_info,"*");
  while (pch != NULL && i < 160)
  {
     str[i]=(char*)malloc(strlen(pch) + 1);
     strcpy(str[i],pch);
     pch = strtok (NULL, "*");
     i++;
  }
total_bukets  = i/8  ;
  for (j=0;str[b_id]!=NULL,j<total_bukets;j++)
  {
  balance_id[j]=str[b_id];
  b_id=b_id+8;
  }
  if (!memcmp(i_use_balance_ind,"Y",1))
  {
     if (atoi(balance_id[0])==1)
     {
        memcpy(o_balance,str[2],16);
        memcpy(o_balance_change,str[3],16);
        memcpy(o_balance_sign,str[7],1);
        for(i=0;i<160;i++)
        free(str[i]);
        return 1;
     }
     else
     {
        for(i=0;i<160;i++)
        free(str[i]);
      return 0;
     }
  }
  else if (!memcmp(i_use_balance_ind,"N",1))
  {
      for (j=1;balance_id[j]!=NULL,j<MAX_BUCKETS;j++)
      {
        b_ind=(j*8)+2;
        bc_ind=(j*8)+3;
        bs_ind=(j*8)+7;
       if (atoi(balance_id[j])!=1 && atoi( str[bc_ind] )!=0)
       {
        memcpy(o_balance,str[b_ind],16);
        memcpy(o_balance_change,str[bc_ind],16);
        memcpy(o_balance_sign,str[bs_ind],1);
        for(i=0;i<160;i++)
        free(str[i]);
        return 1;
       }
      }
     for(i=0;i<160;i++)
     free(str[i]);
    return 0;
  }
 for(i=0;i<160;i++)
 free(str[i]);
return 0;
}

2 个答案:

答案 0 :(得分:1)

我的感觉是这段代码非常脆弱。如果有良好的输入(我不建议桌面检查你的东西),它可能会工作,但如果给出一些不正确的输入,它将崩溃和燃烧或给出误导性的结果。

您是否测试过意外输入?例如:

  • 假设i_balance_info为空?
  • 假设i_balance_info是“”?
  • 假设输入字符串中的项目少于8个,这行代码的作用是什么?

    memcpy(o_balance_sign,str[7],1);
    
  • 假设str [3]中的项目长度少于16个字符,这行代码会做什么?

    memcpy(o_balance_change,str[3],16);
    

我编写此类代码的方法是防范所有此类可能性。至少我会添加ASSERT()语句,我通常会编写显式输入验证并在错误时返回错误。这里的问题是接口似乎不允许存在任何可能输入错误的可能性。

答案 1 :(得分:0)

我很难读你的代码但是FWIW我添加了一些评论,HTH:

// do shorter functions, long functions are harder to follow and make errors harder to spot
// document all your variables, at the very least your function parameters
// also what the function is suppose to do and what it expects as input
int CMI9_auxc_parse_balance_info
(
  char *i_balance_info,
  char *i_use_balance_ind,
  char *o_balance,
  char *o_balance_change,
  char *o_balance_sign
)
{
  char *balance_id[MAX_BUCKETS] = {NULL};
  char balance_info[BALANCE_INFO_FIELD_MAX_LENTH] = {0};
  char *str[160] = {NULL};
  int i=0,j=0,b_id=0,b_ind=0,bc_ind=0,bs_ind=0,rc;
  int total_bukets=0; // good practice to initialize all variables

  //
  // check for null pointers in your arguments, and do sanity checks for any
  // calculations
  // also move variable declarations to just before they are needed
  //

  memset(balance_info,' ',BALANCE_INFO_FIELD_MAX_LENTH);
  memcpy(balance_info,i_balance_info,BALANCE_INFO_FIELD_MAX_LENTH);
  //balance_info[BALANCE_INFO_FIELD_MAX_LENTH]='\0';  // should be BALANCE_INFO_FIELD_MAX_LENTH-1

  char *pch = strtok (balance_info,"*"); // this will potentially crash since no ending \0

  while (pch != NULL && i < 160)
  {
    str[i]=(char*)malloc(strlen(pch) + 1);
    strcpy(str[i],pch);
    pch = strtok (NULL, "*");
    i++;
  }
  total_bukets  = i/8  ;
  // you have declared char*str[160] check if enough b_id < 160
  // asserts are helpful if nothing else assert( b_id < 160 );
  for (j=0;str[b_id]!=NULL,j<total_bukets;j++)
  {
    balance_id[j]=str[b_id];
    b_id=b_id+8;
  }
  // don't use memcmp, if ('y'==i_use_balance_ind[0]) is better
  if (!memcmp(i_use_balance_ind,"Y",1))
  {
    // atoi needs balance_id str to end with \0 has it?
    if (atoi(balance_id[0])==1)
    {
      // length assumptions and memcpy when its only one byte
      memcpy(o_balance,str[2],16);
      memcpy(o_balance_change,str[3],16);
      memcpy(o_balance_sign,str[7],1);
      for(i=0;i<160;i++)
        free(str[i]);
      return 1;
    }
    else
    {
      for(i=0;i<160;i++)
        free(str[i]);
      return 0;
    }
  }
  // if ('N'==i_use_balance_ind[0]) 
  else if (!memcmp(i_use_balance_ind,"N",1))
  {
    // here I get a headache, this looks just at first glance risky. 
    for (j=1;balance_id[j]!=NULL,j<MAX_BUCKETS;j++)
    {
      b_ind=(j*8)+2;
      bc_ind=(j*8)+3;
      bs_ind=(j*8)+7;
      if (atoi(balance_id[j])!=1 && atoi( str[bc_ind] )!=0)
      {
        // length assumptions and memcpy when its only one byte
        // here u assume strlen(str[b_ind])>15 including \0
        memcpy(o_balance,str[b_ind],16);
        // here u assume strlen(str[bc_ind])>15 including \0
        memcpy(o_balance_change,str[bc_ind],16);
        // here, besides length assumption you could use a simple assignment
        // since its one byte
        memcpy(o_balance_sign,str[bs_ind],1);
        // a common practice is to set pointers that are freed to NULL.
        // maybe not necessary here since u return
        for(i=0;i<160;i++)
          free(str[i]);
        return 1;
      }
    }
    // suggestion do one function that frees your pointers to avoid dupl
    for(i=0;i<160;i++)
      free(str[i]);
    return 0;
  }
  for(i=0;i<160;i++)
    free(str[i]);
  return 0;
}

当您想要访问数组中的偏移时,一种有用的技术是创建一个映射内存布局的结构。然后你将指针转换为结构的指针,并使用结构成员来提取信息,而不是你的各种memcpy的

我还建议你重新考虑你的参数到一般的函数,如果你把它们放在一个结构中你有更好的控制,并使函数更具可读性。

int foo( input* inbalance, output* outbalance )

(或者你想要做的任何事情)