c - Access violation when reading array

问题

I am trying to read an array of data and receiving an access violation. I can read the data from the array within the function that the array was allocated using:

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

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

But can't read it outside of the function. This is where I get the access violation, looks like it is trying to read a null location.

How can I access the data in the array AllCurrentData in a different function? thanks!

Extra info:

typedef struct current{

    char **data;

}CurrentData;

AllCurrentData is declared in main:

CurrentData *AllCurrentData = '\0';

Function call

getCurrentData(current, AllCurrentData);

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

回答1:

CurrentData *AllCurrentData = '\0';

This declares a pointer. This pointer is a variables that holds a number that is interpreted as an address. You initialize the pointer to '\0' (null).

getCurrentData(current, AllCurrentData);

Here you pass this pointer as parameter to function getCurrentData. As a pointer is a variable, this variable is passed by value, meaning that the function receives a copy of that value (the number that represents an address).

When inside the function if you write

AllCurrentData = malloc...

you modify that copy of the pointer, so outside the function AllCurrentData will still be '\0'. You need to pass a pointer to that pointer (Inception, I know).

getCurrentData(current, &AllCurrentData);

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

Let me explain this concept on a simpler example:

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.*/
}

so if you want to allocate a pointer inside a function this is not ok:

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.*/
}

The right way to do is:

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] = ...
}

来源：https://stackoverflow.com/questions/20818238/c-access-violation-when-reading-array

标签

pointers

access-violation