Consider the following C code: int interp(int x) f switch(x) f case 6: // BLOCK 1 break; case 5 case 3 // BLOCK 2 break; case 2: case T break; case 8 /BLOCK 4 break; default: // BLOCK 5 // SOME STUFF //SOME STUFF When the code is compiled, the assembly looks like this 01 02 03 04 05 interp: -2(%rdi) , $6 ,%eax $0xc4 *$0x10(,%rax , 8) %eax leal cmp! a. jmpq A friend of yours just debugged this code and told you the starting addresses of the code blocks above BLOCK 1: 0x58 BLOCK 2: 0x80 BLOCK 3: 0x90 BLOCK 4: Oxa8 Based on the information above, answer the questions belovw. For multiple response questions, choose all answers 1. Which descriptions of line 4 in the assembly are correct? (A) The CPU will branch to 0xc4 if x > 6 and do nothing if 0 6 and do nothing if 0