C++ 将 char* 复制到 char*
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copy char* to char*
提问by Zee99
Here is part of my code:
这是我的代码的一部分：
extern "C" REGISTRATION_API int extreme(char* lKey)
{
string s1;
char *p=NULL;
try
{
ifstream myfile ("extreme.txt");
int i=0;
if (myfile.is_open())
{
while (getline(myfile,s1))
{
switch (i)
{
case 1:
strcpy(p,s1.c_str());
lKey=p;
break;
//continue here
}
}
}
}
Now when I call this function from external application, I get this error:
现在，当我从外部应用程序调用此函数时，出现此错误：
AccessViolationException:
Attempted to read or write protected memory. This is often an indication that other memory is corrupt.
AccessViolationException:
试图读取或写入受保护的内存。这通常表明其他内存已损坏。
The problem is due to this:
问题是由于这个：
lKey=p;
How can I assign the lKeyto p?
我怎样才能将 分配lKey给p？
回答by Alexander Gessler
You need to pre-allocate the memory which you pass to strcpy. I.e. a p = new char[s1.length()+1];will do it (+1 for the terminating 0 character). However, it's not a good idea to mix up std::stringand C string routines for no good reason. Better stick with std::string, it will save you a LOTS of trouble.
您需要预先分配您传递给的内存strcpy。即 ap = new char[s1.length()+1];会这样做（+1 表示终止 0 字符）。但是，std::string无缘无故地混用和 C 字符串例程并不是一个好主意。最好坚持使用std::string，它会为您省去很多麻烦。
Also lKey=pwon't work either -- it just copies the local address of pinto the local variable lKey. The caller won't even see a difference.
也lKey=p不会工作——它只是将 的本地地址复制p到本地变量中lKey。来电者甚至看不到任何区别。
回答by James Curran
Actually the problem is strcpy(p,s1.c_str());since pis never set to anything but NULL.
实际上，问题是strcpy(p,s1.c_str());因为p除了 NULL 之外从未设置为任何内容。
回答by Brian R. Bondy
Remember that a char*is just an address of a memory location. You need to have memory allocated at the address.
请记住， achar*只是内存位置的地址。您需要在该地址分配内存。
In your code you don't have memory allocated to use and you didn't set p to point to that memory address.
在您的代码中，您没有分配给使用的内存，也没有将 p 设置为指向该内存地址。
strcpydoes not allocate a buffer, it just takes a memory address to copy the data to.
strcpy不分配缓冲区，只需要一个内存地址来复制数据。
If you are passing a buffer into the function then you probably want simply this (and remove p)
如果您将缓冲区传递给函数，那么您可能只需要这个（并删除 p）
strcpy(lKey, s1.c_str());
回答by jdu.sg
- Eliminate p (it is doing nothing here), and copy the data from s1 directly to lkey
- Not to beat on you, but the indentation scheme is a travesty, please cop a good style from somewhere ( google 1tbs )
- 去掉p（这里什么都不做），直接把s1里面的数据拷贝到lkey
- 不要打败你，但缩进方案是一种讽刺，请从某个地方找到一个好的风格（谷歌 1tbs）
