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中文 (Chinese) / Re: 求写一篇怎么看内盘市场空单怎么成交的简易教程,最好图文说明。
« on: November 29, 2014, 10:54:28 am »
空单基本都只能在price feed筑墙了,而且利率还不能太低。反正我每次看short wall都是几百万BTS。我的理解是这个故意设计成这样好让空单给利息。
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win下客户端编译好了,就是运行报错,出错模块在qtcore.dll上,可能跟版本有关吧,我再想想办法.多谢,我也没有编译过windows版,所以还请多费心。
想看看1s的dpos的稳定性1s不敢试,5s对我都已经是极限了。
你可能最近没有关注Play,Play的重心已经不是博彩平台,而是一个游戏币商城和二手交易所,不过第三方开发者扔可以在其上开发JDS这样的游戏.嗯,我主要是指支持第三方构建可以看做平台。
旗帜鲜明的反对膨胀
这是赤裸裸的掠夺BTS用户
无论用什么口径
且只能给BTC社区一个巨大的笑柄!
3I 可以独断专行
大众只能用脚投票
看看价格吧
就知道!
是否有计划,开发一个博彩平台,可以有双色球,福彩等我的想法还是做个简单的JDS,所以目前没有这个计划。
双色球的结果,可以让代表参考真正双色球的结果发布,51%代表为有效
If we can with the new BitShares on Nov 5th have a some kind of constitution with an unchangeable maximum dilution rate (competitive with Bitcoin), enshrined as the first rule/principle, I personally believe that would be superior to a maximum flexibility DAC.
是否考虑过,运行在bts的块链上运行在bts块链上可能比较小,主要有两个原因:
It is 2/0.8=2,5
That is the limit.. no one will be voted in anywhere near that.
void chain_database_impl::update_active_delegate_list( const full_block& block_data,
const pending_chain_state_ptr& pending_state )
{
std::vector<account_id_type> active_del;
if( block_data.block_num % BTS_BLOCKCHAIN_NUM_DELEGATES == 0 )
{
// perform a random shuffle of the sorted delegate list.
active_del = self->next_round_active_delegates();
} else {
active_del = self->get_active_delegates();
}
auto rand_seed = fc::sha256::hash(self->get_current_random_seed());
uint32_t block_index = block_data.block_num % BTS_BLOCKCHAIN_NUM_DELEGATES;
size_t num_del = active_del.size();
uint32_t signee_delegate_index = -1;
if (block_index!=0) {
//from 0 to where the block signee, we need to swap them to tail to ensure they can only particapant once in a round.
public_key_type block_signee;
block_signee = block_data.signee();
//get delegate's active key
for (uint32_t i=0;i<num_del;++i) {
auto delegate_record = pending_state->get_account_record( active_del[i] );
FC_ASSERT( delegate_record.valid() && delegate_record->is_delegate() );
if (delegate_record->active_key()==block_signee) {
signee_delegate_index = i;
break;
}
}
}
for( uint32_t i = signee_delegate_index+1; i < num_del; ++i )
{
for( uint32_t x = 0; x < 4 && i < num_del; ++x, ++i )
std::swap( active_del[i], active_del[[b]i+(rand_seed._hash[x]%(num_del-i))[/b]] );
rand_seed = fc::sha256::hash(rand_seed);
}
pending_state->set_property( chain_property_enum::active_delegate_list_id, fc::variant(active_del) );
}
You are making sure that a delegate can only participate in a round once on average.It's not 1 delegate in one round but still 50 delegates a round and shuffle the rest delegates after each block.
This means that a delegate does not know in advance when he will go...but when he does get to go he can mine his secrets to make sure he gets to go next....
So you have to wait until at least N delegates have gone before you can be secure. You gain nothing by shuffling every time.
If you want to gain something then you need delegates to commit in advance to 101 secrets to be revealed in time and then shuffle every turn. A delegate would not be able to mine their secret on their turn because they have pre-committed to the next 101 secrets.