分段错误 - 需要帮助解决{新手C ++程序员} [英] Segmentation Fault - Need assistance resolving {Novice C++ Programmer}
问题描述
大家好，我似乎在我的功能中收到了一个段错误但是
不确定如何解决它。我明白这意味着
某处正试图访问未被禁止的内存
或不可用。
以下函数是一个重载的+ =运算符，它以逗号分隔符分隔的记录格式读取
字符串，但
最后一个以分号结尾的数据除外。我将这些记录
添加到存储记录的对象数组中。我希望下面的
代码足以识别问题。静态数组的定义是
，如我的作业所述。
const char * Bank :: operator + =（const char a []）< br $>
{
char stringTemp [316];
char accountTemp [16];
int balanceTemp ;
const char * ptr = a;
sscanf（ptr，"％15 [^，]，％d，％315 [^;] ;"，accountTemp，& balanceTemp，
stringTemp）;
储蓄[size] .changeAccountNumber（accountTemp）;
储蓄[size ] .changeBalance（balanceTemp）;
strcat（stringTemp，" ;;"）;
savings [size] .changeCustomerInfo（stringTemp）;
size ++; //核心转储
返回;
}
AMT2K5写道：大家好，我似乎在我的功能中收到了一个段错误但是不确定如何解决它。我明白这意味着某些地方正试图访问未被禁止或无法使用的内存。
以下函数是一个重载的+ =运算符，它读入
以逗号分隔符分隔的记录格式的字符串，除了以分号结尾的最后一个数据。我将这些记录添加到存储记录的对象数组中。我希望下面的代码足以识别问题。静态数组的定义是在我的任务中说明的。
const char * Bank :: operator + =（const char a []）
{
char stringTemp [316];
确保你使用
char stringTemp [316] = {0};
char accountTemp [16];
在这里相同。
int balanceTemp;
这个还未初始化
const char * ptr = a;
这里没有必要，有吗？你为什么不直接把''a''传递到下面的'/ b $'''扫描片'？
sscanf（ptr，"％15 [^，] ，％d，％315 [^;];"，accountTemp，& balanceTemp，
stringTemp）;
储蓄[size] .changeAccountNumber（accountTemp）;
储蓄[size] .changeBalance（balanceTemp）;
strcat（stringTemp，" ;;"）;
由于''stringTemp''没有初始化为0，它可能会将
分号连接到一些毫无防备的内存部分 - 未定义的行为。 br />
BTW，你确定'stringTemp'中有足够的空间来附加
分号吗？
储蓄[size ] .changeCustomerInfo（stringTemp）;
size ++; //核心转储
返回;
}
V
感谢您的提示。我尝试删除整个功能代码和
留下大小++;但是我仍然有段故障，所以我不确定现在可能是什么问题。
const char * Bank :: operator + = （const char a []）
{
size ++;
}
AMT2K5写道：感谢您的提示。我尝试删除整个功能代码和
留下大小++;但是我仍然有段故障，所以我不确定现在可能出现什么问题。
const char * Bank :: operator + =（const char a []）
{
size ++;
此函数的返回值类型为''const char *''。我没有看到任何
''返回''声明...
这可能是也可能不是您的分段错误的原因。不可能
告诉我们没有看到如何使用这个功能。
}
V
Hello folks, I seem to have recieved a segfault within my function but
am unsure on how to resolve it. I understand that it means that
somewhere something is trying to access memory that is not prohibited
or not available.
The following function is an overloaded += operator that reads in
strings in a record format seperated by comma delimiters except for the
last data which terminates with a semicolon. I am adding these records
to an array of objects which stores the record. I hope the folowing
code is enough to identify the problem. The static array''s are defined
as told in my assignment.
const char* Bank::operator+=(const char a[])
{
char stringTemp[316];
char accountTemp[16];
int balanceTemp;
const char *ptr = a;
sscanf(ptr, "%15[^,],%d,%315[^;];", accountTemp, &balanceTemp,
stringTemp);
savings[size].changeAccountNumber(accountTemp);
savings[size].changeBalance(balanceTemp);
strcat(stringTemp,";");
savings[size].changeCustomerInfo(stringTemp);
size++; //Core dump here
return a;
}
AMT2K5 wrote:Hello folks, I seem to have recieved a segfault within my function but
am unsure on how to resolve it. I understand that it means that
somewhere something is trying to access memory that is not prohibited
or not available.
The following function is an overloaded += operator that reads in
strings in a record format seperated by comma delimiters except for the
last data which terminates with a semicolon. I am adding these records
to an array of objects which stores the record. I hope the folowing
code is enough to identify the problem. The static array''s are defined
as told in my assignment.
const char* Bank::operator+=(const char a[])
{
char stringTemp[316];
Make sure you initiliase it using
char stringTemp[316] = {0};
char accountTemp[16];
Same here.
int balanceTemp;
This one is left uninitialised as well
const char *ptr = a;
There is no need in this, is there? Why don''t you just pass ''a'' to the
''scanf'' below?
sscanf(ptr, "%15[^,],%d,%315[^;];", accountTemp, &balanceTemp,
stringTemp);
savings[size].changeAccountNumber(accountTemp);
savings[size].changeBalance(balanceTemp);
strcat(stringTemp,";");
Since ''stringTemp'' wasn''t initialised to 0 it probably concatenates the
semicolon to some unsuspecting part of memory -- undefined behaviour.
BTW, are you sure there is enough room in ''stringTemp'' to append the
semicolon?
savings[size].changeCustomerInfo(stringTemp);
size++; //Core dump here
return a;
}
V
Thanks for the tips. I tried removing the entire function code and
leaving size++; but I still have a seg fault, so I''m not sure what the
problem may be now.
const char* Bank::operator+=(const char a[])
{
size++;
}
AMT2K5 wrote:Thanks for the tips. I tried removing the entire function code and
leaving size++; but I still have a seg fault, so I''m not sure what the
problem may be now.
const char* Bank::operator+=(const char a[])
{
size++;
This function has the return value type of ''const char*''. I don''t see any
''return'' statement...
That may or may not be the cause of your segmentation fault. Impossible
to tell without seeing how the function is used.
}
V
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