regarding const qualifier in c

[mcbhagyaraj]

sir,
i am getting doubt on const qualifier .
Que:

void main()
{
          const int f(int );
          int b = 4;
          int  a = f(&b);
          printf("%d", a); // o/p i am getting 4 is it ok  or compiler depent and compiled with warnning discarding const qualifier 
 }
 
const int f(int *p)
{     
         *p = 10;
        return  *p; 
}       

compiler : gcc -Linux
Doubt 1: its returning const int and in the main setting in the int variable !
How these is happenning ?
Doubt 2 : int the main , I declared const int a[2] = { 1, 2}; sending f(&a[1])
o/p its reflecting in the main function why?

[weaknessforcats]

First: main()returns an int not a void. main() returning void was an old Microsoft goof-up that they have been trying to eradicate ever since.

Second: This code

void main()
{
const int f(int );
etc...

This is a function prototype:

const int f(int);

because the function returns a copy of an int. This const int can be assigned to either a const or non-const variable:

int x = f(3); //OK

const int y = f(4); //Also OK

so having the return be const is irrelevant. It is ignored by the compiler and also produces a warning.

In general terms, if the function returns a type then having the return be const is ignored and produces a warning.

The above is not true in the case of the function returning a pointer. There the const is important since it describes whether the value pointed at by the pointer is changeable or not. Returning a pointer is a separate case since a pointer is not a type. Rather, it is an address.

Third:

int a = f(&b);

won't compile becuse the function prototype shows an int argument and not an int*.

Fourth:

Doubt 2 : int the main , I declared const int a[2] = { 1, 2}; sending f(&a[1])
o/p its reflecting in the main function why?

You are passing the address of a variable to the function. The function is changing the variable at that address. Hence, you variable in main() got changed.

BTW: Where is const int a[2] = { 1, 2}; ???