I'm confused about the use of the dereference operator in the declaring of char arrays.

I think I understand that the dereference operator is used to dereference memory addresses to their values, and also in an unrelated function in the declaration of pointers; and that arrays are like special pointers to memory addresses.

But why is the * operator used in declaring char arrays, e.g.:

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char * carr = "Hello";

char * carr = "Hello";

Why doesn't the following work, and instead give a series of "invalid conversion from `const char*' to `char'" errors?:

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char carr[5] = {"H", "e", "l", "l", "o"};

char carr[5] = {"H", "e", "l", "l", "o"};

Why are char arrays different than say int arrays?

A char array (and only char arrays), can also be initialised like this (for convenience)
char arr[ ] = "123";
Finally, also for char only, is a pointer to a string constant
char *arr = "123";

A pointer to a string constant is like this to the compiler.

Is there any programming reason why I always see function definitions with arguments like this:

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char getchar(char *arr){

char getchar(char *arr){

Instead of this?:

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char getchar(char arr[]){

char getchar(char arr[]){

Also this code:

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const char anonymous[ ] = "123";
char *arr = anonymous;

const char anonymous[ ] = "123"; char *arr = anonymous;

Didn't work for me, it gives a " invalid conversion from `const char*' to `char*'" compile error.

Would it be correct to say that char *arr = "123"; is like this to the compiler:

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char anonymous[ ] = "123";      
char *arr = anonymous;

char anonymous[ ] = "123"; char *arr = anonymous;

Without the constant declaration?