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sigmentation fault

Discussion in 'C' started by inspire, Aug 24, 2009.

  1. inspire

    inspire New Member

    For the following code, I receive sig fault. Why? Thanks.
    Code:
    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.
     
  2. c_user

    c_user New Member

    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...
     
  3. inspire

    inspire New Member

    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...
     
  4. Amar_Raj

    Amar_Raj New Member

    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.
     
  5. SaswatPadhi

    SaswatPadhi ~ Б0ЯИ Τ0 С0δЭ ~

    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.

    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 :

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

    inspire New Member

    Thank you all!
     

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