我原来在scrypt上做过一个类似想法的原型，供参考。 这个最好配合Layer 1的token方案，不然就要想另外的办法让合约可以验证token的真伪。 生成的测试网交易: https://test.whatsonchain.com/tx/30cc3d743d9e76a314dd7f70aaefabd6bdd976754cd5f9bb8f6c546cb7633f7b contract文件: contract Sell { bytes bsvReceiveScript; int sellAmount; int getAmount; constructor(bytes bsvReceiveScript,int sellAmount, int getAmount) { this.bsvReceiveScript = bsvReceiveScript; this.sellAmount = sellAmount; this.getAmount = getAmount; } public function match(bytes sighashPreimage, bytes receiverScriptPubkey, int coinAmount, int bsvAmount, bytes additionalOutputs) { Tx tx = new Tx(); require(tx.validate(sighashPreimage)); require(coinAmount>0); require(bsvAmount>0); int len = length(sighashPreimage); bytes hashOutputs = sighashPreimage[len - 40 : len - 8]; // scriptCode is just scriptPubKey if there is no CODESEPARATOR in the latter bytes scriptCode = this.readVarint(sighashPreimage[104:]); int scriptLen = length(scriptCode); // last byte contains the state, i.e., counter int amount = unpack(scriptCode[scriptLen - 4 :]); require(coinAmount<=amount); // increment counter bytes scriptCode_ = scriptCode[: scriptLen - 4] ++ num2bin(amount - coinAmount, 4); //todo verify price where require(this.getAmount * coinAmount <= this.sellAmount * bsvAmount); // output: amount + scriptlen + script Sha256 hashOutputs_ = hash256( num2bin(1000, 8) ++ this.writeVarint( b"6a"++b"0453454e44"++this.writeVarint(num2bin(amount - coinAmount, 4)) ++this.writeVarint(num2bin(coinAmount, 4)) ) ++num2bin(1000, 8) ++ this.writeVarint(scriptCode_) ++num2bin(1000, 8) ++ this.writeVarint(receiverScriptPubkey) ++num2bin(bsvAmount, 8) ++ this.writeVarint(this.bsvReceiveScript) ++additionalOutputs ); // ensure output is expected: amount is same with specified // also output script is the same with scriptCode except counter incremented require(hashOutputs == hashOutputs_); } function readVarint(bytes b) returns (bytes) { int len = 0; bytes ret = b""; bytes header = b[0:1]; if (header == b"fd") { len = this.fromLEUnsigned(b[1:3]); ret = b[3:3+len]; } else if (header == b"fe") { len = this.fromLEUnsigned(b[1:5]); ret = b[5:5+len]; } else if (header == b"ff") { len = this.fromLEUnsigned(b[1:9]); ret = b[9:9+len]; } else { len = this.fromLEUnsigned(b[0:1]); ret = b[1:1+len]; } return ret; } function writeVarint(bytes b) returns (bytes) { int n = length(b); bytes header = b""; if (n < 0xfd) { header = this.toLEUnsigned(n, 1); } else if (n < 0x10000) { header = b"fd" ++ this.toLEUnsigned(n, 2); } else if (n < 0x100000000) { header = b"fe" ++ this.toLEUnsigned(n, 4); } else if (n < 0x10000000000000000) { header = b"ff" ++ this.toLEUnsigned(n, 8); } return header ++ b; } // convert signed integer `n` to unsigned integer of `len` bytes, in little endian function toLEUnsigned(int n, int len) returns (bytes) { // one extra byte to accommodate possible negative sign byte bytes m = num2bin(n, len + 1); // remove sign byte return m[0 : length(m) - 1]; } function fromLEUnsigned(bytes b) returns (int) { // append positive sign byte. This does not hurt even when sign bit is already positive return unpack(b ++ b"00"); } }

my question is how to mitigate the address problem. if you want to ship peer to peer with address anonymity, how does international postal system handle delivery with mailer not knowing purchasers address?