What's the difference, in terms of bit structure, between an int and a uint under the EVM?
For an example, if an int8's value, expressed in binary, was 00010011, what would the uint8 of that be? What would negative 00010011 look like?
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Related: ethereum.stackexchange.com/questions/12397/… – Richard Horrocks Jul 14 '17 at 9:02
Edition (Thanks Tjaden) : when an int is negative it is represented using two's complement system in which we we get for an int: 1111 1111=−1(255 for uint) whereas 0000 0001=1 (1 for uint).
For example:
uint256 u=123456789;
is represented by 0x00000000000000000000000000000000000000000000000000000000075bcd15 ( Big Endian)
and int v=-123456789;
is represented by 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffff8a432eb ( Big Endian )
int w=123456789
is represented by :
"0x00000000000000000000000000000000000000000000000000000000075bcd15"
answered Jul 14 '17 at 8:26
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3This is not entirely correct. Solidity (and the signed EVM opcodes) expect signed integers to be in two's complement form. You can't just flip the highest bit and make the number negative – Tjaden Hess♦ Jul 19 '17 at 10:24
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Elaborating on @TjadenHess, you could use the signextend opCode to do this for you. – Vignesh Karthikeyan Jun 4 '19 at 20:29
