[inspire]

For the following code, I receive sig fault. Why? Thanks.

void test(int *i) {
  *i = 3; /* where the sig fault occurs */
}

int main() {
  int *p;
  test(p);
}

BTW, I'm using gcc 4.3.3 on Ubuntu 9.04.

[c_user]

see friend as per my logic u cannot assign a no. to a pointer so its showing u the fault....
>>*i=3 is the wrong statement....
hav a gud day...

[inspire]

actually, I wasn't assigning a number to a point. *i is dereferencing a pointer. *i = 3 is to let i point to value 3. I'm still confused why this caused the seg fault.

This is a simplified version of the problem. Originally, I wanted to manipulate a string parameter in a function. It seems as long as I change the value pointed by a pointer (this pointer is a parameter to a function), seg fault appears...

[Amar_Raj]

Since it is not initialized with valid address and you are accessing that address to modify the value, the segmentation fault occurs! Just allocate some valid address and try... it should work.

[SaswatPadhi]

Quote:

Originally Posted by c_user

see friend as per my logic u cannot assign a no. to a pointer so its showing u the fault....
>>*i=3 is the wrong statement....
hav a gud day...

He is not assigning the int to the pointer. He is actually de-referencing the the pointer with *, and assigning 3 to the address to which the pointer is pointing.

Quote:

Originally Posted by Amar_Raj

Since it is not initialized with valid address and you are accessing that address to modify the value, the segmentation fault occurs! Just allocate some valid address and try... it should work.

Right.
You are de-referencing a pointer which is not assigned to any address.

It's always a good practice to check the pointer b4 assigning values to it. Like this :

void test(int *i) {
  if(i != NULL) { *i = 3; }
  else              { printf("ERROR :: Pointer does not point to a valid address !\n"); }
}

[inspire]

Thank you all!