Question

Consider the below 2 declarations.

* appears next to the datatype and not next to variable

char* ptr1, * ptr2, * ptr3; //all 3 are pointers

* appears next to the variable and not next to datatype

char *ptr1,*ptr2,*ptr3; //again all 3 are pointers

Is there any difference in intepretation between the 2 declarations. I know there is no difference in the variables.

What is the rationale behind introducing void pointers?

Answer 1

The only difference is whitespace, they end up being pointers-to-chars. I personally use char * ptr1, using plenty of whitespace in my code.

Void pointers can point to any type of object and have no size. For example, this works (but isn’t really a good idea):

int * ptr1;
float * ptr2;
(*ptr1) = 17;
ptr2 = (void*)ptr1;

(*ptr2) will now interpret the bytes of ptr1 as a float instead of an int, giving you a different result. You can change between types. There are some uses for void*s in pointer-to-function typedefs, as well.

In addition, if you have char * ptr1 and do ptr1++, you will increment the address ptr1 refers to by sizeof(char). That’s not possible with a void*, it will give you a compiler error saying it doesn’t know what size to increment by.