将char *复制到char * [英] copy char* to char*

问题描述

这是我的代码的一部分：

Here is part of my code:

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
         }
      }
   }
}

现在，当我从外部应用程序调用此函数时，会收到以下错误：

Now when I call this function from external application, I get this error:

AccessViolationException：

尝试读取或写入受保护的内存。这通常表示其他内存已损坏。

AccessViolationException:
Attempted to read or write protected memory. This is often an indication that other memory is corrupt.

问题是由于以下原因：

lKey=p;

如何将 lKey code> p ？

How can I assign the lKey to p?

推荐答案

您需要预先分配您传递的内存到 strcpy 。也就是说a p = new char [s1.length（）+ 1]; 将执行它（终止0字符为+1）。但是，混合 std :: string 和C字符串例程并不是一个好主意。更好地使用 std :: string ，它会为你节省很多麻烦。

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::string and C string routines for no good reason. Better stick with std::string, it will save you a LOTS of trouble.

也 lKey = p 不会工作 - 它只是将 p 的本地地址复制到本地变量 lKey 。呼叫者甚至不会看到差异。

Also lKey=p won't work either -- it just copies the local address of p into the local variable lKey. The caller won't even see a difference.

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