Set - 5

Question 1 :

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

Answer :

In the first case 6 bytes are allocated to the variable a which is fixed, where as in the second case if *p is assigned to some other value the allocate memory can change.


Question 2 :

What's the auto keyword good for?

Answer :

Answer1
Not much. It declares an object with automatic storage duration. Which means the object will be destroyed at the end of the objects scope. All variables in functions that are not declared as static and not dynamically allocated have automatic storage duration by default. 

For example

int main(){
	int a; //this is the same as writing "auto int a;"
} 

Answer2
Local variables occur within a scope; they are "local" to a function. They are often called automatic variables because they automatically come into being when the scope is entered and automatically go away when the scope closes. The keyword auto makes this explicit, but local variables default to auto auto auto auto so it is never necessary to declare something as an auto auto auto auto.


Question 3 :

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

Answer :

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)……


Question 4 :

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

Answer :

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.


Question 5 :

What does extern mean in a function declaration?

Answer :

It tells the compiler that a variable or a function exists, even if the compiler hasn't yet seen it in the file currently being compiled. This variable or function may be defined in another file or further down in the current file.