Let's break down the bytecode and understand the flow of execution to find the answer:
-
CALLVALUE (34)
: This opcode gets the value sent with the transaction and pushes it onto the stack. -
CODESIZE (38)
: This opcode pushes the size of the code (in bytes) onto the stack. -
SUB (03)
: This opcode pops two values from the stack, subtracts the second one from the first, and pushes the result back onto the stack. -
JUMP (56)
: This opcode alters the program counter to the location specified by the top stack item. It jumps to the location if it's a validJUMPDEST
. -
JUMPDEST (5B)
: Marks a valid destination for jumps. -
STOP (00)
: Halts execution.
- At byte 00,
CALLVALUE
pushes a value (let's call itV
) onto the stack. - At byte 01,
CODESIZE
pushes the size of the code onto the stack. Since the code is 10 bytes long, the value 10 (in hexadecimal, '0A') is pushed onto the stack. - At byte 02,
SUB
subtracts the top two values on the stack. The stack has0A
(fromCODESIZE
) andV
(fromCALLVALUE
). The operation is0A - V
, and the result is pushed back onto the stack. - At byte 03,
JUMP
uses the result ofSUB
to determine where to jump. For a valid jump, this result must match the address of aJUMPDEST
. In this code, the onlyJUMPDEST
is at byte 06.
To find the answer, we need to determine the value of V
such that 0A - V
equals 06
. This calculation is straightforward:
0A - V = 06 (hex)
10 - V = 06 (decimal)
V = 10 - 06
V = 04
The answer is 4 (in decimal) or 04 (in hexadecimal), as this would be the value pushed by CALLVALUE
that allows the JUMP
to successfully target the JUMPDEST
at byte 06.