(from older bitcoin code: bitcoin-0.5.1-win32-original):
main.cpp
int64 static GetBlockValue(int nHeight, int64 nFees)
{
int64 nSubsidy = 50 * COIN;
// Subsidy is cut in half every 4 years
nSubsidy >>= (nHeight / 210000);
return nSubsidy + nFees;
}
(from newest bitcoin code: 1 march 2014):
int64_t GetBlockValue(int nHeight, int64_t nFees)
{
int64_t nSubsidy = 50 * COIN;
// Subsidy is cut in half every 210,000 blocks which will occur approximately every 4 years.
nSubsidy >>= (nHeight / Params().SubsidyHalvingInterval());
return nSubsidy + nFees;
}
So code seems to be something like:
Reward = Reward >> SomeValue; // de-obfuscated
What will happen depends on how this will be compiled to instructions and how those instructions behave.
For now x86 and x64/amd64 is popular so a good guess is intel's instruction manual.
Two possible instructions are likely chosen by a C compiler:
1. shld
2. shr
shld may produce undefined results so the reward could be even higher than the default.
however shr is probably chosen by compilers, and shld will probably also behave as if the shift value was masked.
And that is what happens for both instructions, the shift value is masked by 5 bits or 6 bits all being 1's depending on if it's 32 bit or 64 bit mode.
However I think bitcoin is probably compiled to run on 32 bit operating systems as well, which might mean that shld was used or perhaps some compiler specific 64 bit software implementation of shr.
One would have to look at the instructions generated for this specific function to be sure what's going to happen... but since it was tested... we kinda know.
Theoretically at least the shifter is wrapping back as follows:
64 mod 64 = 0 so no shift happens.
Which is the same as
64 and 111111b
This explains the reward of 50 * COIN
So to calculate when this wrap around of the shift value would happen requires this to be solved:
nHeight / 210000 = 64
which leads to:
nHeight = 64 * 210000
which gives:
13440000
^ the block count/number mentioned.
nHeight is probably the block number and each block is generated every 10 minutes.
The 210k is not exactly 4 years so that a bit sloppy.
Anyway one year is: 365.25 * 24 * 60 / 10 = 52596 blocks.
So this block number will be reached from genesis at:
13440000 / 52596 = 255.5 years.
4 years is: 210384
Perhaps 210k was chosen a bit sloppy to offset that blocks are found a bit faster than 10 minutes, probably not the reason though... just sloppy.
In reality this block number will be reached a bit faster probably because of investments in new hardware and so forth... though nobody can predict the future lol.
Now for some interesting questions... how to fix this ?
There are supposed to be only 21 million coins or something... so that means:
21.000.000 / 50 = 420000 blocks. though... the number of blocks will be much higher because rewards will be cut in half and so forth... each 4 years...
So this makes it a bit more difficult but still kinda easy to calculate:
First 210k blocks: 50 coins
Second 210k blocks: 25 coins
Third 210k blocks: 12.5 coins
Fourth 210k blocks: 6.25 coins
Fiveth 210k blocks: 3.125 coins
Sixth 210k blocks: 1.5625 coins.
So let's start:
210.000 x 50 = 21000000 - 10500000 = 10500000
210.000 x 25 = 10500000 - 5250000 = 5250000
210.000 x 12.5 = 5250000 - 2625000 = 2625000
210.000 x 6.25 = 2625000 - 1312500 = 1312500
and so forth...
So what we need to calculate now is the number of times this 21m number can be halved.
Which is log10(21m) / log10(2) = 24.323885992102934376117624838116
So that's roughly:
24.32 * 210k blocks = 5108016.0583416162189847012160044 blocks.
5.1m blocks will be mined and then the rewards should go to zero.
Putting this back into the function gives:
5108016.0583416162189847012160044 / 210000 = 24.323885992102934376117624838116
The shift value of 24 is therefore well in range of the 0 to 31 or 63 bit value range.
However... just because no more rewards are given out doesn't actually stop the block number from going up...
blocks will still be mined just no more rewards given... so that is indeed a bit of a problem.
After 255 years... a bit of a lol there... suddenly rewards would be given out which would be funny... or perhaps even unspecified ammounts of rewards.
Anyway a simple sloppy fix could be something like:
if nHeight <= 5109017 then
// calculate reward
else
// no reward.
I am not sure about that fractional part
So just to be fair to whoever mines it last... perhaps one extra block for you... perhaps max will be slightly over 21m or slightly under depending on what you guys would want.
if it must stay under 21m then:
if nHeight <= 5109016 then
// calculate reward
else
// no reward.
So from this little inspection there does seem to be a little problem with bitcoin... it won't actually produce 21m bitcoins exactly... but something near it... unless it's somehow solved to give that tiny little fraction to the last block or so... or perhaps it will fit perfect by chance/rounding me not sure... would be nice if someone investigates that further just for the fun of it
However the future is not certain... perhaps 10 or 20 years from now intel is dead and this code might behave totally different
Or how about this for something really cool:
if nHeight <= 5109016 then
// calculate reward
if nHeight = 5109017 then
// calculate bitcoins remaining of that 21m and give it to the last block, risky but cool
(if calculation fucks up that would be one wealthy block
)
// probably safest to perform this calculation offline and just put a constant here or so
else
// no reward.
I also wonder if old clients that may be used by then would be exploiteable... perhaps they would assume rewards were given out... perhaps there are other checks in bitcoin or perhaps not...
Would be fun for those future programmers/hackers to exploit old clients like that bihihi... some old clients would then maybe accept payments from those blocks
Perhaps some bitcoin vending machines/atms/hardware.. though commercial interests would probably update those... though bitcoin hardware still a good guess... stuffed away somewhere in some bank
Bye,
Skybuck.
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