c++ - 指向预分配内存的指针作为输入参数并让函数填充它

Question

测试代码:

void modify_it(char * mystuff)
{
   //last element is null i presume for c style strings here.
   char test[7] = "123456";

   //when i do this i thought i should be able to gain access to this
   //bit of memory when the function is destroyed but that does not
   //seem to be the case.
   //static char test[] = "123123";

   //this is also creating memory on stack and not the heap i reckon
   //and gets destroyed once the function is done with.
   //char * test = new char[7];

   //this does the job as long as memory for mystuff has been
   //allocated outside the function.
   strcpy_s(mystuff,7,test); 

   //this does not work. I know with c style strings you can't just do
   //string assignments they have to be actually copied. in this case
   //I was using this in conjunction with static char test thinking
   //by having it as static the memory would not get destroyed and i can
   //then simply point mystuff to test and be done with it. i would later
   //have address the memory cleanup in the main function.
   //but anyway this never worked.
   mystuff = test;
}

int main(void)
{
  //allocate memory on heap where the pointer will point 
  char * mystuff = new char [7];
  modify_it(mystuff);
  std::string test_case(mystuff);

  //this is the only way i know how to use cout by making it into a c++ string.
  std::cout<<test_case.c_str();

  delete [] mystuff;
  return 0;
}

1. 对于函数中的静态数组，为什么它不起作用？
2. 如果我在函数中使用 new 分配内存，它是在堆栈还是堆上创建的？
3. 如果我有一个需要复制到 char * 形式的字符串。我看到的所有内容通常都需要 const char* 而不仅仅是 char*。

我知道我可以使用引用来轻松解决这个问题。或者 char ** 发送指针并以这种方式执行。但我只是想知道我是否可以仅使用 char * 来做到这一点。无论如何，您的想法和评论以及任何示例都会非常有帮助。

Answer 1

char * mystuff = new char [7];
delete mystuff;

delete mystuff 导致未定义的行为。您必须删除[]您新[]的内容。