Just a guess, but assuming the value is being treated as a signed 32 bit integer the highest value that can be stored would be 2147483647. Add one to that and the computer will see it as -2147483648.
The number you gave is so large it will actually wrap around several times. My binary math is really rusty but I if this the case then as far the trader is concerned you are offering him a 251,635,157 deficit in trading value. He is understandably unwilling to take a 250 million buck loss on a trade.
Of course this whole post begs the question of what exactly is worth over 200 billion dorf bucks? Glitched artifact?
For reference: 214,496,729,643 in binary is 11000111110001000000000101101000101011. Only the last 32 bits will be kept because the number is too large. Leaving us with 11110001000000000101101000101011. A signed integer sees the last 1 as an indication that the number is negative. To get the positive value of the same magnitude we must take the complement of that binary number (all 0s become 1s, and 1s become 0s) and add 1.
11110001000000000101101000101011
xor 11111111111111111111111111111111
---------------------------------------------
= 00001110111111111010010111010100
+ 00000000000000000000000000000001
---------------------------------------------
= 00001110111111111010010111010101
In decimal that's 251,635,157. (so the computer sees it as -251,635,157)
Author
Topic: I was trading... (Read 505 times)