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. -
DUP1 (80)
: Duplicates the top value of the stack. -
MUL (02)
: Multiplies the top two values on the stack and pushes the result back onto the stack. -
PUSH2 0100
: Pushes the value 0100 (256 in decimal) onto the stack. -
EQ (14)
: Compares the top two values on the stack for equality. If equals, it pushes 1 (true) onto the stack; otherwise, it pushes 0 (false). -
PUSH1 0C
: Pushes the value 0C (12 in decimal) onto the stack. -
JUMPI (57)
: If the top value of the stack is true (1), it alters the program counter to the value beneath it (in this case, 0C), performing a conditional jump to the JUMPDEST at byte 0C. -
JUMPDEST (5B)
: Marks a valid destination for jumps. -
STOP (00)
: Halts execution.
For the JUMPI
at byte 09
to jump to the JUMPDEST
at byte 0C
, the result of the EQ
operation must be true
.
-
CALLVALUE: Let's call the value to send as
X
-
DUP1: Stack now has two
X
-
MUL:
X
*X
-
PUSH2 0100:
0100
is pushed to stack, or 256 (in decimal) -
EQ: (because next set of op-code is
JUMPI
)X * X = 256 (X)^2 = 256 X = 16