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?

Let's trace the execution of the program step by step:

1. Global variables:a = 0, b = 0, c = 0 (uninitialized global a and b default to 0).

2. main() begins:

   • static int a = 1; initializes a static variable a local to main with the value $1$.

   • First call to prtFun():

     • static int a = 2; initializes static local a in prtFun to $2$.

     • int b = 1; initializes auto local b in prtFun to $1$.

     • a += ++b; increments b from $1$ to $2$, then a becomes $2 + 2 = 4$.

     • printf prints: 4 2.

   • Back in main():

     • a += 1; increments main's static variable a from $1$ to $2$.

   • Second call to prtFun():

     • static int a retains its value from the previous call ($4$).

     • int b = 1; re-initializes auto local b to $1$.

     • a += ++b; increments b from $1$ to $2$, then a becomes $4 + 2 = 6$.

     • printf prints: 6 2.

   • Back in main():

     • `printf("

%d %d ", a, b);`: - a refers to main's static local variable, which has the value $2$. - Since there is no local variable b in main(), it refers to the global variable b, which is $0$. - It prints: 2 0.

Thus, the complete printed output is: ``` 4 2 6 2 2 0 ```

Therefore, option (C) is the correct choice.