c - 释放对象的问题

Question

我目前正在尝试学习 C++ 的 C 方面。

我尝试为 256 的 char 数组malloc 一 block 内存，然后我给它分配了一个 char* "Hello World!" 但是当我释放对象时出现错误。

任何人都可以向我解释错误。

#include <exception>
#include <stdexcept>
#include <iostream>

int main()
{
    void* charVoidPointer = malloc( sizeof(char) * 256 ) ;
    charVoidPointer = "Hello World";

    std::cout << (char *)charVoidPointer;
    free (charVoidPointer);
}

Answer 1

“Hello World”由编译器静态分配。它是程序的一部分，存在于程序可寻址的某个地方；称之为地址 12。

charVoidPointer 最初指向某个由 malloc 为您分配的位置；称它为地址 98。

charVoidPointer = "Hello ..."使 charVoidPointer 指向您程序中的数据；地址 12。您丢失了先前包含在 charVoidPointer 中的地址 98。

而且你不能释放不是由 malloc 分配的内存。

为了更真实地展示我的意思:

void* charVoidPointer = malloc(sizeof(char) * 256);
printf("the address of the memory allocated for us: %p\n", charVoidPointer);
charVoidPointer = "Hello World";
printf("no longer the address allocated for us; free will fail: %p\n",
       charVoidPointer);

你的意思是:

strcpy(charVoidPointer, "Hello World");

编辑:其他类型的寻址内存示例

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int main()
{
  // an array of 10 int
  int *p = (int*)malloc(sizeof(int) * 10);

  // setting element 0 using memcpy (works for everything)
  int src = 2;
  memcpy(p+0, &src, sizeof(int));

  // setting element 1 using array subscripts.  correctly adjusts for
  // size of element BECAUSE p is an int*.  We would have to consider
  // the size of the underlying data if it were a void*.
  p[1] = 3;

  // again, the +1 math works because we've given the compiler 
  // information about the underlying type.  void* wouldn't have
  // the correct information and the p+1 wouldn't yield the result
  // you expect.
  printf("%d, %d\n", p[0], *(p+1));

  free (p);
}

实验；将类型从 int 更改为 long、double 或一些复杂类型。