10

Yes, it's golf. So sue me.

Rules

The response to getCode() should have some characters that aren’t either 0 or x.

To test it out, you can use the following with a local testrpc:

web3.eth.sendTransaction({data:'BYTECODE', from:web3.eth.accounts[0]});
web3.eth.getCode('THE_ADDRESS');

Or you can use the evm command that comes with Geth:

evm --debug --code BYTECODE run.

Example:

This is a valid, but long-ish example: 0x600180600b6000396000f3

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4 Answers 4

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3

Since Noel got close, time to release my 6-byter:

0x3859818153F3

Bytecode:

CODESIZE MSIZE DUP2 DUP2 MSTORE8 RETURN

Works as follows:

Use CODESIZE to get a nonzero value (0x06) on the stack using only one byte. MSIZE puts 0 on the stack. This will deploy the contract 0x060000000000.

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3
  • So, you were behind this all along? :D
    – Noel Maersk
    Feb 24, 2018 at 13:57
  • No, no - this is @maurelian's thing, but I did get a little head start. I'm reasonably sure that 6 bytes is minimal, but somebody surprise me!
    – benjaminion
    Feb 24, 2018 at 14:04
  • Well, MSTORE and RETURN will consume 4 stack entries, so it makes sense that you'd need those 2, plus 4 more to load the stack. Unless there's a way to place two at once?
    – maurelian
    Feb 25, 2018 at 14:44
5

6 bytes (code: 6 bytes, payload: 0 bytes); 3 distinct opcodes. Same idea just 2 gas cheaper than previous 6-byter.

Bytecode:

0x3838533838f3

Assembly:

CODESIZE CODESIZE MSTORE8 CODESIZE CODESIZE RETURN

Works as follows:

Use the cheapest single-byte operation to get a value onto the stack CODESIZE. Store CODESIZE bytes at CODESIZE address in memory and then return CODESIZE bytes from CODESIZE location in memory.

Sample transaction: on etherscan

Gas used to deploy: 54622.

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3
  • That "payload" in the template might have been misleading. What I meant by "payload" was data-as-code that's customisable, i.e. the actual value in this many bytes does not strictly change the bytecode's operation, and can be used to modify the target contract. (This is an unnecessary feature under the exercise's rules.)
    – Noel Maersk
    Feb 24, 2018 at 17:53
  • @NoelMaersk - in that case, my solution would have a payload of 0 because you cannot customize the payload at all. Am I understanding that correctl?
    – yarrumretep
    Feb 25, 2018 at 15:55
  • Yup, and that's perfectly fine! :)
    – Noel Maersk
    Feb 25, 2018 at 19:51
1

7 bytes (code: 6 bytes, payload: 1 byte); 4 distinct opcodes.

Bytecode:

0x600180808053f3

Assembly:

PUSH1 0x01 DUP1 DUP1 DUP1 BYTES8 RETURN

Works as follows:

Push a byte of value X to the stack. Duplicate it three times. Move a byte from stack to memory, to same location X as the byte's value. Return X bytes from memory, starting at location X.

Sample transaction: on etherscan or etherchain.

Gas used to deploy: 53694.

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0

Meta: template used for my answer:

X bytes (code: X bytes, payload: X byte); X distinct opcodes.

Bytecode:

>! 0xXXXXXXXXXXXXX

Assembly:

>! XXXXXXXXXXXXXXX

Works as follows:

>! XXXXXXXXXXXXXXX
>! XXXXXXXXXXXXXXX
>! XXXXXXXXXXXXXXX

Sample transaction: on [etherscan](XXX) or [etherchain](XXX).

Gas used to deploy: XXXXX.
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