difference between char *(arr[5]) and char (*arr)[5]

Question

It appears that I don't practically know the difference between char *(arr[5]) and char (*arr)[5] Logically, as it is with type-definitions, char *(arr[5]) is a pointer to an array of 5 char 's, char *arr[5] is an array of char pointers then what (*arr)[5] will be then? Or it makes no difference ?

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I did some tests and found out that these 3 declarations are different. For instance if I use char (*arr)[5] it totally allows me to do that arr = malloc(2) , otherwise it moans about illegal conversions.

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Can someone explain the differences between these declarations, when they are used and it will be a huge plus if this also includes may I do arr = malloc(2) to dynamically allocate an array of [5] chars, as it allows me to semantically do so with char (*arr)[5] although giving me stack overflows when being used. (disclaimer: [] suggests stack allocation, so it doesn't make much sense as a heap memory address cannot point to stack memory address as far as I am not aware)

Answer 1

char *(arr[5]) equals to char *arr[5] and it means you have array of pointers that each of them points to a character so arr[0] can points to a character and arr[1] can points to a character and so on. For example you can write your code like this:

char *(arr[5]);// or char *arr[5]
char a = 'a',b='b',c='c';
arr[0] = &a;
arr[1] = &b;
arr[2] = &c;    
printf("a=%c b=%c c=%c \n",*arr[0],*arr[1],*arr[2]);

But (*arr)[5] means you have an array of 5 characters and you want to point it.
For example you can write your code like this:

char (*arr)[5];
char str[5] = "Hell";
arr = &str;
printf("%s\n",*arr);