I want to get an estimation of how much it would cost to store arbitrarily many elements in a mapping of the form BYTES32 -> BYTES32. I see here that a STORAGEADD operation costs 20000 gas, which I believe is the instruction to add an element to a mapping (is it correct?).

The gas price today is 20 gwei, so 20 * 10^9 wei which gives me a total of 20000 * 20 * 10^9 * 10^-18 = 0.0004 ether ~ 0.0048 USD (with 1 ETH = 12 USD).

Is it correct and is there any limit to how many elements I can put into the mapping (disregarding gas price limitations)?

2 Answers 2


The total storage is made of 32-bytes slots addressed with 256 bits. This gives us 2^256 * 32 bytes to use. When you add an item to a mapping, it is sent to a random location in the storage calculated by sha3, see this answer.

Adding an item to a mapping will never fail because there always will be a sha3-calculated location to put the info into. Of course, the closer you get to 2^256 insertions, the higher the likelihood that you will eventually overwrite something else.

  • 1
    A good rule of thumb is that you need approximately sqrt(n) entries before you expect a collision, so 2^128 in this case. That's about 3.4e38, which is 340 million million million million million million. en.wikipedia.org/wiki/Birthday_problem#Square_approximation There are also means to correct for hash collisions in maps, allowing for theoretically infinite entries, but I'm not sure of the exact implementation in the EVM. Commented Jul 3, 2021 at 13:45

See this response to a similar question. In short: you can store any number of elements in a mapping.

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