To interact with any level you will send raw bytes over stdin to this program.

To efficiently solve these problems, first run it to see the challenge instructions.

Then craft, assemble, and pipe your bytes to this program.

For instance, if you write your assembly code in the file asm.S$,$ you can assemble that to an object file:

as $-$o asm.o asm.S

Note that if you want to use Intel syntax for x$86($which$,$ of course, you do$),$ you'll need to add the following to the start of asm.S:

$.$intel$\_$syntax noprefix

Then, you can copy the $.$text section $($your code$)$ to the file asm.bin:

objcopy $-$O binary $--$only$-$section$=.$text asm.o asm.bin

And finally, send that to the challenge:

cat $./$asm$.$bin $|/$challenge$/$run

You can even run this as one command:

as $-$o asm.o asm.S && objcopy $-$O binary $--$only$-$section$=.$text $./$asm$.$o $./$asm$.$bin && cat $./$asm$.$bin $|/$challenge$/$run

In this level you will be working with control flow manipulation. This involves using instructions

to both indirectly and directly control the special register $`$rip$`,$ the instruction pointer.

You will use instructions such as: jmp$,$ call, cmp$,$ and their alternatives to implement the requested behavior.

We will be testing your code multiple times in this level with dynamic values! This means we will

be running your code in a variety of random ways to verify that the logic is robust enough to

survive normal use.

The last jump type is the indirect jump, which is often used for switch statements in the real world.

Switch statements are a special case of if$-$statements that use only numbers to determine where the control flow will go$.$

Here is an example:

switch$($number$)$:

$0$: jmp do$\_$thing$\_0$

$1$: jmp do$\_$thing$\_1$

$2$: jmp do$\_$thing$\_2$

default: jmp do$\_$default$\_$thing

The switch in this example is working on $`$number$`,$ which can either be $0,1,$ or $2.$

In the case that $`$number$`$ is not one of those numbers, the default triggers.

You can consider this a reduced else$-$if type structure.

In x$86,$ you are already used to using numbers, so it should be no suprise that you can make if statements based on something being an exact number.

In addition, if you know the range of the numbers, a switch statement works very well.

Take for instance the existence of a jump table.

A jump table is a contiguous section of memory that holds addresses of places to jump.

In the above example, the jump table could look like:

$[0$x$1337]=$ address of do$\_$thing$\_0$

$[0$x$1337+0$x$8]=$ address of do$\_$thing$\_1$

$[0$x$1337+0$x$10]=$ address of do$\_$thing$\_2$

$[0$x$1337+0$x$18]=$ address of do$\_$default$\_$thing

Using the jump table, we can greatly reduce the amount of cmps we use.

Now all we need to check is if $`$number$`$ is greater than $2.$

If it is$,$ always do:

jmp $[0$x$1337+0$x$18]$

Otherwise:

jmp $[$jump$\_$table$\_$address $+$ number $*8]$

Using the above knowledge, implement the following logic:

if rdi is $0$:

jmp $0$x$403014$

else if rdi is $1$:

jmp $0$x$4030$f$3$

else if rdi is $2$:

jmp $0$x$40319$d

else if rdi is $3$:

jmp $0$x$40326$e

else:

jmp $0$x$40336$a

Please do the above with the following constraints:

Assume rdi will NOT be negative

Use no more than $1$ cmp instruction

Use no more than $3$ jumps $($of any variant$)$

We will provide you with the number to 'switch' on in rdi.

We will provide you with a jump table base address in rsi.

Here is an example table:

$[0$x$40435$b$]=0$x$403014($addrs will change$)$

$[0$x$404363]=0$x$4030$f$3$

$[0$x$40436$b$]=0$x$40319$d

$[0$x$404373]=0$x$40326$e

$[0$x$40437$b$]=0$x$40336$a

Please give me your assembly in bytes $($up to $0$x$1000$ bytes$)$: