GO Classes 2023 | Weekly Quiz 6 | Question: 12

Answer 1

we’ll call putStar( , ) as PS( , )

str[] = “hello”; strlen(str) = 5

1. PS(0,str) → upStar[0] = str[0] = ‘h’; 0 != 5 thus upStar[1] = ‘*’; nextcall
2. PS(1,str) → upStar[1] = str[1] = ‘e’; 1 != 5 thus upStar[2] = ‘*’; nextcall
3. PS(2,str) → upStar[2] = str[2] = ‘l’; 2 != 5 thus upStar[3] = ‘*’; nextcall
4. PS(3,str) → upStar[3] = str[3] = ‘l’; 3 != 5 thus upStar[4] = ‘*’; nextcall
5. PS(4,str) → upStar[4] = str[4] = ‘o’; 4 != 5 thus upStar[5] = ‘*’; nextcall
6. PS(5,str) → upStar[5] = str[5] = ‘\0’; 5 == 5 thus return;

Thus, upStar[] contains “hello\0”. Notice, every ‘*’ is again overwritten by corresponding char from str[].

Answer :- D. hello