[英]How to perform a deep copy of a char *?
我對如何執行char *的深拷貝a感到有些困惑。 這就是我所擁有的:
Appointment(Appointment& a)
{
subject = new char;
*subject = *(a.subject);
}
Appointment(Appointment& b)
{
location = new char;
*location = *(b.location);
}
char *subject;
char *location;
我正在嘗試對char指針的主題和位置進行深層復制。 這樣行嗎? 如果不是,那么有關如何執行此操作的任何建議?
由於您使用的是C ++,因此應使用std::string
滿足您的字符串需求。
您編寫的以下代碼
Appointment(Appointment& a)
{
subject = new char;
*subject = *(a.subject);
}
不會做您想做的事,在上面分配一個字符( new char
),然后將a.subject
的第一個字符分配給它( *subject = *(a.subject)
)
為了復制char*
指向的字符串,您必須首先確定字符串長度,分配內存以容納該字符串,然后復制字符。
Appointment(Appointment& a)
{
size_t len = strlen(a.subject)+1;
subject = new char [len]; // allocate for string and ending \0
strcpy_s(subject,len,a.subject);
}
char*
的另一種方法是使用std::vector<char>
,這取決於您要對字符串執行的操作。
您必須跟蹤char*
長度才能復制它們,例如:
class Appointment
{
public:
char *subject;
int subject_len;
char *location;
int location_len;
Appointment() :
subject(NULL), subject_len(0),
location(NULL), location_len(0)
{
}
~Appointment()
{
delete[] subject;
delete[] location;
}
Appointment(const Appointment& src) :
subject(new char[src.subject_len]), subject_len(src.subject_len),
location(new char[src.location_len]), location_len(src.location_len)
{
std::copy(src.subject, src.subject + src.subject_len, subject);
std::copy(src.location, src.location + src.location_len, location);
}
Appointment& operator=(const Appointment& lhs)
{
delete[] subject;
subject = NULL;
delete[] location;
location = NULL;
subject = new char[lhs.subject_len];
subject_len = lhs.subject_len;
std::copy(lhs.subject, lhs.subject + lhs.subject_len, subject);
location = new char[lhs.location_len];
location_len = lhs.location_len;
std::copy(lhs.location, lhs.location + lhs.location_len, location);
}
};
在這種情況下,最好使用std::string
代替:
class Appointment
{
public:
std::string subject;
std::string location;
Appointment()
{
}
Appointment(const Appointment& src) :
subject(src.subject), location(src.location)
{
}
Appointment& operator=(const Appointment& lhs)
{
subject = lhs.subject;
location = lhs.location;
}
};
這可以進一步簡化,因為由編譯器生成的默認構造函數和賦值運算符足以自動為您深度復制值:
class Appointment
{
public:
std::string subject;
std::string location;
};
沒有。
您需要分配足夠的內存來存儲要復制的字符串
subject = new char [strlen(a.subject + 1]; // +1 to allow for null charcter terminating the string.
然后使用strncpy
memcpy
或在循環中復制所有字符以復制字符串
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