C double character pointer declaration and initialization

Question

I always though that declaring

char *c = "line"; 

was the same as

char c[] = "line"; 

and so I did

char **choices = { "New Game", "Continue Game", "Exit" }; 

Which gives me an incompatible pointer type, where

char *choices[] = { "New Game", "Continue Game", "Exit" }; 

doesn't. Any help on understanding this?

Answer 1

char *c = "line"; 

is not the same as

char c[] = "line"; 

it's really the same as

static const char hidden_C0[] = "line"; char *c = (char *)hidden_C0; 

except that the variable hidden_C0 is not directly accessible. But you'll see it if you dump out generated assembly language (it will usually have a name that isn't a valid C identifier, like .LC0). And in your array-of-string-constants example, the same thing is going on:

char *choices[] = { "New Game", "Continue Game", "Exit" }; 

becomes

const char hidden_C0[] = "New Game"; const char hidden_C1[] = "Continue Game"; const char hidden_C2[] = "Exit";  char *choices[] = { (char *)hidden_C0, (char *)hidden_C1, (char *)hidden_C2 }; 

Now, this is a special case behavior that is available only for string constants. You cannot write

int *numbers = { 1, 2, 3 }; 

you must write

int numbers[] = { 1, 2, 3 }; 

and that's why you can't write

char **choices = { "a", "b", "c" }; 

either.

(Your confusion is a special case of the common misconception that arrays are "the same as" pointers in C. They are not. Arrays are arrays. Variables with array types suffer type decay to a pointer type when they are used (in almost every context), but not when they are defined.)