C - char array and char pointer -

why can't define array

char **pp={ "123", "456", "789" }; 

but can define char*[] ,and send function accept char **

char *pp[]={ "123", "456", "789" }; fun(pp);  void fun(char **pointertopointer) {     //++(**pointertopointer);//error     printf("%s", *pointertopointer);  } //output::"123" 

and why can't increment

++(**pointertopointer); 

to answer first question, principles might clearer if use single depth of pointer. code illegal same reason:

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

in c, braced initializer list not object (especially not array). can taken list of items read initializers when object being initialized.

ptr 1 object, @ 1 initializer taken, , expected form of initializer pointer (which 1 not).

in fact code explicitly illegal under c11 6.7.9/11:

the initializer scalar shall single expression, optionally enclosed in braces

however, there gcc bug/feature permits excess initializers scalar , ignores them. further, compilers may "be helpful" , "only" issue warning, , initialize ptr point address 1, wherever might be.

"scalar" means object that's not struct or array.

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since c99 can write:

int *ptr = (int []){1, 2, 3}; 

which creates array (using same storage duration ptr) , points ptr @ first element.

this array mutable; non-mutable 1 use int const *ptr = (int const[]){1, 2, 3}; instead.

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replacing int char *, see write:

char **p = (char *[]){ "123", "456", "789" }; 

in case the pointers in array mutable, things point (i.e. string literals) still aren't.

note should use char const * when dealing string literals, because not mutable. fact string literals have type char [n] historical hangover before const added c. so:

char const **pp = (char const *[]){ "123", "456", "789" }; 

or non-mutable pointers strings:

char const *const *pp = (char const *const []){ "123", "456", "789" };