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;
}
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)