I am new to Solidity, so please, accept my apologies, if the question is really dumb. I am trying to write two functions, one of which gets the uint256 and converts it to a binary number, and the other gets the binary input (if that's even possible) and returns a uint256. Is such a conversion even possible?

  • Can you describe more of what you mean by a "binary number?" Do you mean a string representation of binary (e.g. "00000111" to represent 7)? Or do you mean converting between uint256 and bytes32? Or something else? – user19510 Feb 13 '18 at 3:25
  • I meant converting a 7 00111 (bytes16 I guess). But if this is not possible to do to get a number, I can go with a string representation, with the ability of how to convert it back to a number as an output. – Ruham Feb 13 '18 at 3:32
  • I'm still not sure what you mean. bytes16 is a sequence of 16 bytes of data. Those bytes can be anything and can be interpreted any way. (You could use it to store a number of a string.) Maybe you can give an example of the input and the output you want or further explain what you're trying to accomplish. – user19510 Feb 13 '18 at 4:14
  • Another way to maybe help with the confusion... what is a "binary number" to you? (In a computer, all numbers are internally stored as binary, but that's rarely relevant when you write code. x = 7 involves binary in the CPU, but why do you care?) – user19510 Feb 13 '18 at 4:15
  • Thanks for clarifying, I'll try to explain what I mean. I want a function, that takes a two-digit number, and converts it into a binary representation of 5 binary numbers. For example, the input would be 11, and the output would be 01101. Then, I want another function to reverse the input of 01101 to output as 11, if that's possible to do. – Ruham Feb 13 '18 at 4:47

I still can't imagine a use for this, but I wrote code to do it because it seemed fun:

pragma solidity ^0.4.19;

contract Convert {
    function toBinaryString(uint8 n) public pure returns (string) {
        // revert on out of range input
        require(n < 32);

        bytes memory output = new bytes(5);

        for (uint8 i = 0; i < 5; i++) {
            output[4 - i] = (n % 2 == 1) ? byte("1") : byte("0");
            n /= 2;

        return string(output);

    function fromBinaryString(bytes5 input) public pure returns (uint8) {
        uint8 n = 0;

        for (uint8 i = 0; i < 5; i++) {
            n *= 2;
            if (input[i] == "1") {
                n += 1;
            } else {
                // revert on malformed input
                require(input[i] == "0");

        return n;
  • The reason would be, if you have a set numbers representing different flag states, such as 1, 2, 4,, 8, 16, and you wanted to identify which ones were in your sample number, the simple way would be to Xor your number. but alas you can not do that with an integer. – Cyberience Mar 27 '19 at 3:34
  • I'm not sure what you're saying. You can certainly xor two integers in Solidity. (Although in the case of checking for flags, don't you want and? E.g. if (flags & 1) ..., if (flags & 2) ...) – user19510 Mar 27 '19 at 3:36
  • I am working on something similar and tried your suggestion before you replied, and using solidity 0.5.0 in remix I get the TypeError: Type uint8 is not implicitly convertible to expected type bool. if (_restrictionCode & 8) { – Cyberience Mar 27 '19 at 3:48
  • 1
    Sorry, try if (_restrictionCode & 8 != 0). – user19510 Mar 27 '19 at 3:55

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