giving a char* an address gets me "does not name a type"

Isn't this the proper way to init a pointer?

char *event[6];
int e = 6;
event = &e;

It errors out on 'event' does not name a type

One of the correct variants is

int *event;
int e = 6;
event = &e;

But what exactly you are trying to achieve?

alesam: One of the correct variants is

int *event;
int e = 6;
event = &e;

But what exactly you are trying to achieve?

difficult to guess without a code.

char *event[6]; declares an array of pointers, perfectly legal. Is that what you want?
Later you can use this to store the addresses of characters → event[2] = &some_char_variable;

In your example, if you want event (which is undefined) to hold the address of e, you either need to declare event as a pointer ( char *event = &e;) or use one of your array positions (event[0] = &e;).
Note that using the same identifier for each of these in the same code space is not permitted.

mattlogue: Isn't this the proper way to init a pointer?

char *event[6];
int e = 6;
event = &e;

It errors out on 'event' does not name a type

iowevent[1] = &e; would be legal. You are trying to assign a value to a variable which has not been previously declared.

mattlogue: Isn't this the proper way to init a pointer?

char *event[6];
int e = 6;
event = &e;

It errors out on 'event' does not name a type

That is not the error my compiler shows.

Arduino: 1.8.9 (Windows 10), Board: "Arduino/Genuino Uno"

C:\Users\XXXX\AppData\Local\Temp\arduino_modified_sketch_125381\sketch_apr09a.ino: In function 'void setup()':

sketch_apr09a:7:7: error: incompatible types in assignment of 'int*' to 'char* [6]'

 event = &e;

       ^

exit status 1
incompatible types in assignment of 'int*' to 'char* [6]'

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