比特币源码阅读——货币供应

1. 说明

比特币的初始货币供应是对问题解决者成功计算的 50 亿聪的奖励。每 210,000 个区块（10 分钟/区块大约需要 4 年）后，奖励呈几何级数减少：

求和公式计算总供应量（为什么 n=32比特币的源代码在哪里，因为 32 的奖励是 1 satoshi，然后 /2 奖励就没有了）：

源码实现：

在比特币源代码中，总量（2100万亿聪）不是直接设定的，而是从每210,000个区块减半奖励中推断出来的

相关源代码：

class CMainParams : public CChainParams:
# 创世区块的50亿聪比特币的奖励 (50 * COIN)
genesis = CreateGenesisBlock(1231006505, 2083236893, 0x1d00ffff, 1, 50 * COIN);
# 减半周期是210000个块, 210000.0 * 10 / 60 / 24 /365 = 4  (大约4年减半一次)
consensus.nSubsidyHalvingInterval = 210000;
# 减半的逻辑
GetBlockSubsidy

总计算：

from decimal import *
COIN = 10 ** 8        
base_reward = 50 * COIN
adjust_section = 210000
all_sectio_num = 33
def get_decimal_str(num):
        return "{0:.8f}".format(((Decimal(num) / COIN).quantize(Decimal('0.00000000'))))
        
sum = 0
     
for i in range (0, all_sectio_num):
    nSubsidy = base_reward
    nSubsidy = nSubsidy >> i
    sum += adjust_section * nSubsidy
print("sum: %s" % sum)
print("sum decimal: %s" % get_decimal_str(sum))

分阶段打印：

from decimal import *
sum = 2099999997690000
COIN = 10 ** 8        
base_reward = 50 * COIN
adjust_section = 210000
all_sectio_num = 33 + 1
def get_decimal_str(num):
        return "{0:.8f}".format(((Decimal(num) / COIN).quantize(Decimal('0.00000000'))))
        
def percentage(a, b):
    if (abs(1.0 * b - 0.0) < 0.000001):
        return "infinite"
    else:
        return ("%.8f%%" % (1.0 * a / b * 100))
summined = 0
print("s s s s s s s s" % ("Block", "Reward Era", "BTC/block", "Start BTC", "BTC Added", "End BTC", "BTC Increase", "End BTC % of Limit"))
for i in range(0 , all_sectio_num):
    block = i * adjust_section
    
    nSubsidy = base_reward
    nSubsidy = nSubsidy >> i
    currentmined = adjust_section * nSubsidy 
    endedbtc = summined + currentmined
    
    increase = percentage(currentmined, summined)
    endedbtcoflimit = percentage(endedbtc, sum)
    print("d d s s s s s s" %(block, i+1, get_decimal_str(nSubsidy), get_decimal_str(summined), get_decimal_str(currentmined), get_decimal_str(endedbtc), increase, endedbtcoflimit))
    summined = summined + currentmined

理论时间计算（以实际为准）：

比特币减半是由明确的 210,000 个区块决定的。而每 4 年减半的奖励也只是从这个前提推导出来的。根据以下脚本比特币的源代码在哪里，如果挖矿算力不变，大约有 2140 个比特币将被挖出

from decimal import *
import datetime
sum = 2099999997690000
COIN = 10 ** 8        
base_reward = 50 * COIN
adjust_section = 210000
all_sectio_num = 33 + 1
beginepoch = 1231006505 # GMT+08:00: 2009年1月4日星期日凌晨2点15分
oneyear = 60 * 60 * 24 * 365 
def get_decimal_str(num):
        return "{0:.8f}".format(((Decimal(num) / COIN).quantize(Decimal('0.00000000'))))
        
def percentage(a, b):
    if (abs(1.0 * b - 0.0) < 0.000001):
        return "infinite"
    else:
        return ("%.8f%%" % (1.0 * a / b * 100))
# GMT+08:00
def getepochstr(epoch):
    return datetime.datetime.fromtimestamp(epoch).strftime('%Y-%m-%d')
current_epoch = beginepoch
summined = 0
print("s s s s s s s s s s" % ("Date reached", "Block","Reward Era", "BTC/block", "Year (estimate)", "Start BTC", "BTC Added", "End BTC", "BTC Increase", "End BTC % of Limit"))
for i in range (0, all_sectio_num):
    # 每一轮打印4年
    
    for j in range(0, 4):
        block = i * adjust_section + adjust_section/4 * j
        nSubsidy = base_reward
        nSubsidy = nSubsidy >> i
        currentmined = adjust_section / 4  * nSubsidy 
        endedbtc = summined + currentmined
        
        increase = percentage(currentmined, summined)
        endedbtcoflimit = percentage(endedbtc, sum)
        
        estimate = current_epoch + oneyear
        print("s d d s s s s s s s" % (getepochstr(current_epoch), block, i+1, get_decimal_str(nSubsidy), getepochstr(estimate), get_decimal_str(summined), get_decimal_str(currentmined), get_decimal_str(endedbtc),increase, endedbtcoflimit))
        current_epoch = current_epoch + oneyear
        summined = summined + currentmined

2. 参考资料