What is the unsigned range of a bytes32 type in Solidity?

I'm wondering the total number of unique expressible combinations. I'm assuming it is 2^8^32.

  • 1
    You are correct bytes32 have 32 bytes you have 256^32 = (2^8)^32 possible combinations. – Ismael Oct 8 '17 at 17:16
  • @Ismael Can you add this as an answer so I can mark it as answered? – Alec Kriebel Oct 9 '17 at 16:40

You are correct bytes32 has 32 bytes, each byte has 256 combination then you will have 256×256×..{32 veces}..×256 = 256^32 = (2^8)^32 possible combinations total.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.