Question

我正在尝试读取数据数组并接收访问冲突。我可以使用以下命令从数组中读取数组中的数据：

AllCurrentData[newLineCount].data[tabCount] = malloc(length + 1);
strcpy( AllCurrentData[newLineCount].data[tabCount], buffer );

printf("%s", AllCurrentData[newLineCount].data[tabCount]);

但是在功能之外无法读取它。这是我获得访问冲突的地方，看起来它正在尝试读取空位置。

如何在不同的函数中访问数组AllCurrentData中的数据？谢谢！

额外信息：

typedef struct current{

    char **data;

}CurrentData;

AllCurrentData在main中声明：

CurrentData *AllCurrentData = '\0';

函数调用

getCurrentData(current, AllCurrentData);

printf("%s", AllCurrentData[0].data[0]);   //<----- error here

Answer 1

CurrentData *AllCurrentData = '\0';

这声明了一个指针。该指针是一个变量，它包含一个被解释为地址的数字。将指针初始化为'\ 0'（null）。

getCurrentData(current, AllCurrentData);

这里将此指针作为参数传递给函数getCurrentData。由于指针是变量，因此该变量通过值传递，这意味着该函数接收该值的副本（表示地址的数字）。

如果你写了

，在函数内部

AllCurrentData = malloc...

你修改了指针的副本，所以在函数AllCurrentData之外仍然是'\ 0'。你需要传递一个指向那个指针的指针（我知道它是一开始的。）

getCurrentData(current, &AllCurrentData);

getCurrentData(.., CurrentData **p_AllCurrentData) {
*p_AllCurrentData = malloc(...);
}

让我在一个更简单的例子中解释这个概念：

int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address)
v = malloc(10 * sizeof(int)); // here the pointer is assigned a valid address, lets say 0x41A0
f(v);


void f(int *x) {
  // here the pointer x receives a copy of the value of the pointer v, so x will be, like v, 0x41A0
  x[0] = 4; 
  /*this is ok as you modify the memory at the address 0x41A0,
  so this modification is seen from outside the function, as v points to the same address.*/

  x = malloc(...);
  /* this is NOT ok, as x receives a new address, lets say 0xCC12.
  But because x has a copy of the value from v, v will still be unmodified.*/
}

所以如果你想在函数内部分配一个指针，那就不行了：

int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address)

f(v);    

void f(int *x) {
  // here the pointer x receives a copy of the value of the pointer v, so x will be, like v, NULL


  x = malloc(...); 
  /*this is NOT ok, as x receives a new address, lets say 0xCC12.
  But because x has a copy of the value from v, v will still be unmodified,
  it will still have NULL.*/
}

正确的方法是：

int *v = NULL; // here v is a pointer. This pointer has value 0 (invalid address)
// as v is a variable (of type pointer to int), v has an address, lets say 0xAAAA.
f(&v);


void f(int **p_x) {
  /* here the p_x is a pointer to (a pointer of int),
  so this pointer receives a copy of the value the address of p, so p_x is 0xAAAA.*/


  *p_x = malloc(...);
  /* lets say malloc returns the address 0xBBBB. This will be the address of our vector.
  *p_x= says we modify the value at the address p_x. So the value found at the adress 0xAAAA
  will be 0XBBBB. But because 0xAAAA is the address of v, we effectively modified the value of v
  to 0xBBBB. So now v has the address of our starting vector.*/
  // if you want to modify values inside this vector you need:
  (*p_x)[0] = ...
}

c - 读取数组时的访问冲突

1 个答案: