虽然 @Burt 的答案看起来是正确的(虽然没有测试),但有一种更有效的方法来解决相同的任务:
function toHex16 (bytes16 data) internal pure returns (bytes32 result) {
result = bytes32 (data) & 0xFFFFFFFFFFFFFFFF000000000000000000000000000000000000000000000000 |
(bytes32 (data) & 0x0000000000000000FFFFFFFFFFFFFFFF00000000000000000000000000000000) >> 64;
result = result & 0xFFFFFFFF000000000000000000000000FFFFFFFF000000000000000000000000 |
(result & 0x00000000FFFFFFFF000000000000000000000000FFFFFFFF0000000000000000) >> 32;
result = result & 0xFFFF000000000000FFFF000000000000FFFF000000000000FFFF000000000000 |
(result & 0x0000FFFF000000000000FFFF000000000000FFFF000000000000FFFF00000000) >> 16;
result = result & 0xFF000000FF000000FF000000FF000000FF000000FF000000FF000000FF000000 |
(result & 0x00FF000000FF000000FF000000FF000000FF000000FF000000FF000000FF0000) >> 8;
result = (result & 0xF000F000F000F000F000F000F000F000F000F000F000F000F000F000F000F000) >> 4 |
(result & 0x0F000F000F000F000F000F000F000F000F000F000F000F000F000F000F000F00) >> 8;
result = bytes32 (0x3030303030303030303030303030303030303030303030303030303030303030 +
uint256 (result) +
(uint256 (result) + 0x0606060606060606060606060606060606060606060606060606060606060606 >> 4 &
0x0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F) * 7);
}
function toHex (bytes32 data) public pure returns (string memory) {
return string (abi.encodePacked ("0x", toHex16 (bytes16 (data)), toHex16 (bytes16 (data << 128))));
}
此代码产生大写输出。对于小写输出,只需将代码中的 7 更改为 39 即可。
解释
这个想法是使用二进制运算一次处理 16 个字节。
The toHex16
函数将 16 个字节的序列转换为bytes16
值转换为 32 个十六进制数字的序列,表示为bytes32
价值。这toHex
函数拆分abytes32
值一分为二bytes16
块,通过以下方式将每个块转换为十六进制表示toHex16
函数,最后连接0x
使用转换后的块作为前缀abi.encodePacked
功能。
最复杂的部分是如何toHex16
功能有效。我们逐句解释一下。
第一句话:
result = bytes32 (data) & 0xFFFFFFFFFFFFFFFF000000000000000000000000000000000000000000000000 |
(bytes32 (data) & 0x0000000000000000FFFFFFFFFFFFFFFF00000000000000000000000000000000) >> 64;
这里我们将输入的最后 64 位右移 64 位,基本上是:
0123456789abcdeffedcba9876543210
\______________/\______________/
| |
| +---------------+
______V_______ ______V_______
/ \ / \
0123456789abcdef0000000000000000fedcba9876543210
第二句:
result = result & 0xFFFFFFFF000000000000000000000000FFFFFFFF000000000000000000000000 |
(result & 0x00000000FFFFFFFF000000000000000000000000FFFFFFFF0000000000000000) >> 32;
这里我们将两个 64 位块的最后 32 位右移 32 位:
0123456789abcdef0000000000000000fedcba9876543210
\______/\______/ \______/\______/
| | | |
| +-------+ | +-------+
__V___ __V___ __V___ __V___
/ \ / \ / \ / \
012345670000000089abcdef00000000fedcba980000000076543210
下一句话:
result = result & 0xFFFF000000000000FFFF000000000000FFFF000000000000FFFF000000000000 |
(result & 0x0000FFFF000000000000FFFF000000000000FFFF000000000000FFFF00000000) >> 16;
does:
012345670000000089abcdef00000000fedcba980000000076543210
\__/\__/ \__/\__/ \__/\__/ \__/\__/
| | | | | | | |
| +---+ | +---+ | +---+ | +---+
V_ V_ V_ V_ V_ V_ V_ V_
/ \ / \ / \ / \ / \ / \ / \ / \
012300004567000089ab0000cdef0000fedc0000ba980000765400003210
下一个:
result = result & 0xFF000000FF000000FF000000FF000000FF000000FF000000FF000000FF000000 |
(result & 0x00FF000000FF000000FF000000FF000000FF000000FF000000FF000000FF0000) >> 8;
does:
012300004567000089ab0000cdef0000fedc0000ba980000765400003210
\/\/ \/\/ \/\/ \/\/ \/\/ \/\/ \/\/ \/\/
| | | | | | | | | | | | | | | |
| +-+ | +-+ | +-+ | +-+ | +-+ | +-+ | +-+ | +-+
V V V V V V V V V V V V V V V V
/\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\
01002300450067008900ab00cd00ef00fe00dc00ba00980076005400320010
本系列的最后一句话有点不同:
result = (result & 0xF000F000F000F000F000F000F000F000F000F000F000F000F000F000F000F000) >> 4 |
(result & 0x0F000F000F000F000F000F000F000F000F000F000F000F000F000F000F000F00) >> 8;
它将奇数半字节右移 4 位,偶数半字节右移 8 位:
01002300450067008900ab00cd00ef00fe00dc00ba00980076005400320010
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
V V V V V V V V V V V V V V V V V V V V V V V V V V V V V V V V
000102030405060708090a0b0c0d0e0f0f0e0d0c0b0a09080706050403020100
因此,初始数据的所有半字节都分布在每个字节中。
现在每个字节x
我们需要做如下改造:
x` = x < 10 ? '0' + x : 'A' + (x - 10)
让我们稍微重写一下这个公式:
x` = ('0' + x) + (x < 10 ? 0 : 'A' - '0' - 10)
x` = ('0' + x) + (x < 10 ? 0 : 1) * ('A' - '0' - 10)
注意(x < 10 ? 0 : 1)
可以计算为((x + 6) >> 4)
,因此我们有:
x` = ('0' + x) + ((x + 6) >> 4) * ('A' - '0' - 10)
x` = (0x30 + x) + ((x + 0x06) >> 4) * 7
最后声明:
result = bytes32 (0x3030303030303030303030303030303030303030303030303030303030303030 +
uint256 (result) +
(uint256 (result) + 0x0606060606060606060606060606060606060606060606060606060606060606 >> 4 &
0x0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F) * 7);
基本上对每个字节都执行上述计算。这
0x0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F0F
需要右移后的掩码将原始公式中右移“丢弃”的位清零。
顺便说一句,最好在以下位置提出这样的问题:https://ethereum.stackexchange.com/