After some careful analysis, you believe that exactly one instruction is corrupted.

Somewhere in the program, either a jmp is supposed to be a nop, or a nop is supposed to be a jmp. (No acc instructions were harmed in the corruption of this boot code.)

The program is supposed to terminate by attempting to execute an instruction immediately after the last instruction in the file. By changing exactly one jmp or nop, you can repair the boot code and make it terminate correctly.

For example, consider the same program from above:

nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6

If you change the first instruction from nop +0 to jmp +0, it would create a single-instruction infinite loop, never leaving that instruction. If you change almost any of the jmp instructions, the program will still eventually find another jmp instruction and loop forever.

However, if you change the second-to-last instruction (from jmp -4 to nop -4), the program terminates! The instructions are visited in this order:

nop +0  | 1
acc +1  | 2
jmp +4  | 3
acc +3  |
jmp -3  |
acc -99 |
acc +1  | 4
nop -4  | 5
acc +6  | 6

After the last instruction (acc +6), the program terminates by attempting to run the instruction below the last instruction in the file. With this change, after the program terminates, the accumulator contains the value 8 (acc +1, acc +1, acc +6).

Assignment

Fix the program so that it terminates normally by changing exactly one jmp (to nop) or nop (to jmp). What is the value of the accumulator after the program terminates? Determine this in the following way:

• Write a function fix that takes the pathname (char*) of a text file containing the boot code of a game console. The function must return the value (int) of the accumulator after the given code has been fixed so that it terminates.

Example

In this interactive session we assume the text file code.txt¹ to be located in the current directory.

> fix("code.txt")
8