Why can't I directly assign an int to an int pointer like this: int *p = 6;?

Question

error: invalid conversion from 'int' to 'int*'
int *q = 8;

Works fine.
*q = 6;

Why can't I directly assign an int to an int pointer like this: int *q = 6; and I can assign it safely in the next line?

Answer 1

Because they're different things at all. The 1st one is definition of variable with initializer expression, i.e an initialization (of the pointer itself):

int *         q                      = 8;
~~~~~         ~                      ~~~
type is int*; name of variable is q; initialized with 8

The 2nd one is assignment (of the object pointed by the pointer):

*q                              = 6;
~~                              ~~~
dereference on q via operator*; assign the resulting lvalue pointed by q to 6

And, int *p = 6; means define a variable named p with type int* and initialize it with 6, which fails because 6 can't be used for initializing a pointer directly (i.e. the error "invalid conversion from 'int' to 'int*'").

Answer 2

* symbol is reused for two different purpuses in your snippet. First time it is used as a part of type declaration int *, declaring a pointer to int. Second time it is used to dereference a pointer *q, invoking indirection operator.

* can also be used to invoke multiplication operator *q = *q * *q;;

To assign a value to an integer pointed by pointer you need to dereference it. And to assigning integral value other than 0 to a pointer itself (that is what int *q = 8; is doing) requires reinterpret_cast, hence you get this error.