Declaring type of pointers?

Question

I just read that we need to give the type of pointers while declaring them in C (or C++) i.e.:

int *point ;

As far as I know, pointers sto

Answer 1

One reason is in pointer arithmetic. You cannot do p+1 unless you know the size of the element to which p points -- that is the size of the type to which p is a pointer. If you'd try p+1 on a void *p you're likely to get a bad answer (it is the same as if done on a char * but maybe you didn't want that; it is caught by -pedantic as a warning and by -pedantic-errors as an error).

Another reason is type safety. If a function receives as argument an int * you cannot pass a pointer to char (a string) there. You'd get a warning (an error with -Werror / -pedantic-errors). Consider this (dummy) code:

void test(int *x)
{
}

int main()
{
    char *x = "xyz";
    test(x);
    return 0;
}

Compiling (using gcc (GCC) 4.8.2 20131017 (Red Hat 4.8.2-1)) gives:

1.c: In function ‘main’:
1.c:8:2: warning: passing argument 1 of ‘test’ from incompatible pointer type [enabled by default]
  test(x);
  ^
1.c:1:6: note: expected ‘int *’ but argument is of type ‘char *’
 void test(int *x)
      ^