Consider the following C code segment.int a, b, c = 0; void prtFun(void); main( ) { static int a = 1; /* Line 1 */ prtFun( ); a += 1; prtFun( ); printf(" %d %d ", a, b); } void prtFun(void) { static int a = 2; /* Line 2 */ int b = 1; a += ++b; printf(" %d %d ", a, b); }What output will be generated by the given code segment if: Line 1 is replaced by auto int a = 1; Line 2 is replaced by register int a = 2;

Replacing Line 1 and Line 2 changes the storage classes:

• Line 1: auto int a = 1; (an automatic variable local to main).

• Line 2: register int a = 2; (an automatic/register variable local to prtFun, re-initialized on every invocation).

Let's trace the execution:

1. main() begins:

   • auto int a = 1; initializes main's local a to $1$.

   • First call to prtFun():

     • register int a = 2; initializes a to $2$.

     • int b = 1; initializes b to $1$.

     • a += ++b; increments b to $2$, and a becomes $2 + 2 = 4$.

     • printf prints: 4 2.

   • Back in main():

     • a += 1; increments main's local a to $1 + 1 = 2$.

   • Second call to prtFun():

     • register int a = 2; re-initializes a to $2$ (since register variables have automatic storage duration and do not retain values across calls).

     • int b = 1; initializes b to $1$.

     • a += ++b; increments b to $2$, and a becomes $2 + 2 = 4$.

     • printf prints: 4 2.

   • Back in main():

     • `printf("

%d %d ", a, b); prints main's local a ($2$) and the global b` ($0$). - It prints: 2 0.

Thus, the complete output is: ``` 4 2 4 2 2 0 ```

Therefore, option (D) is the correct choice.