What is the difference between char a[] = “string”; and char *p = “string”; ?

Answer1

a[] = “string”;
char *p = “string”;

The difference is this:

p is pointing to a constant string, you can never safely say
p[3]=’x';
however you can always say a[3]=’x';
char a[]=”string”; - character array initialization.
char *p=”string” ; - non-const pointer to a const-string.( this is permitted only in the case of char pointer in C++ to preserve backward compatibility with C.)

Answer2

a[] = “string”;
char *p = “string”;
a[] will have 7 bytes. However, p is only 4 bytes. P is pointing to an adress is either BSS or the data section (depending on which compiler — GNU for the former and CC for the latter).

Answer3

char a[] = “string”;
char *p = “string”;

for char a[]…….using the array notation 7 bytes of storage in the static memory block are taken up, one for each character and one for the terminating nul character.
But, in the pointer notation char *p………….the same 7 bytes required, plus N bytes to store the pointer variable “p” (where N depends on the system but is usually a minimum of 2 bytes and can be 4 or more)……

How do I declare an array of N pointers to functions returning pointers to functions returning pointers to characters?

Answer1

If you want the code to be even slightly readable, you will use typedefs.
typedef char* (*functiontype_one)(void);
typedef functiontype_one (*functiontype_two)(void);
functiontype_two myarray[N]; //assuming N is a const integral

Answer2

char* (* (*a[N])())()
Here a is that array. And according to question no function will not take any parameter value.