Erlang io：将二进制文件格式化为十六进制
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【中文标题】Erlang io：将二进制文件格式化为十六进制【英文标题】：Erlang io:formatting a binary to hex 【发布时间】：2011-04-15 15:43:33 【问题描述】：我可以格式化 Erlang 二进制文件，以便每个字节都以十六进制写入吗？即，
> io:format(???, [<<255, 16>>]).
<<FF, 10>>
我在io:format 文档中没有看到明显的方法，但也许我只是缺少一个？将二进制文件转换为列表并分别格式化其元素效率太低了。
【问题讨论】：
【参考方案1】：不，没有这样的格式化选项，但您可以执行以下操作：
io:format("<<~s>>~n", [[io_lib:format("~2.16.0B",[X]) || <<X:8>> <= <<255,16>> ]]).
如果您需要，还有更快的解决方案。
-module(bin_to_hex).
-compile([native, hipe, [o3]]).
-export([bin_to_hex/1]).
bin_to_hex(B) when is_binary(B) ->
bin_to_hex(B, <<>>).
-define(H(X), (hex(X)):16).
bin_to_hex(<<>>, Acc) -> Acc;
bin_to_hex(Bin, Acc) when byte_size(Bin) band 7 =:= 0 ->
bin_to_hex_(Bin, Acc);
bin_to_hex(<<X:8, Rest/binary>>, Acc) ->
bin_to_hex(Rest, <<Acc/binary, ?H(X)>>).
bin_to_hex_(<<>>, Acc) -> Acc;
bin_to_hex_(<<A:8, B:8, C:8, D:8, E:8, F:8, G:8, H:8, Rest/binary>>, Acc) ->
bin_to_hex_(
Rest,
<<Acc/binary,
?H(A), ?H(B), ?H(C), ?H(D), ?H(E), ?H(F), ?H(G), ?H(H)>>).
-compile(inline, [hex/1]).
hex(X) ->
element(
X+1, 16#3030, 16#3031, 16#3032, 16#3033, 16#3034, 16#3035, 16#3036,
16#3037, 16#3038, 16#3039, 16#3041, 16#3042, 16#3043, 16#3044,
16#3045, 16#3046, 16#3130, 16#3131, 16#3132, 16#3133, 16#3134,
16#3135, 16#3136, 16#3137, 16#3138, 16#3139, 16#3141, 16#3142,
16#3143, 16#3144, 16#3145, 16#3146, 16#3230, 16#3231, 16#3232,
16#3233, 16#3234, 16#3235, 16#3236, 16#3237, 16#3238, 16#3239,
16#3241, 16#3242, 16#3243, 16#3244, 16#3245, 16#3246, 16#3330,
16#3331, 16#3332, 16#3333, 16#3334, 16#3335, 16#3336, 16#3337,
16#3338, 16#3339, 16#3341, 16#3342, 16#3343, 16#3344, 16#3345,
16#3346, 16#3430, 16#3431, 16#3432, 16#3433, 16#3434, 16#3435,
16#3436, 16#3437, 16#3438, 16#3439, 16#3441, 16#3442, 16#3443,
16#3444, 16#3445, 16#3446, 16#3530, 16#3531, 16#3532, 16#3533,
16#3534, 16#3535, 16#3536, 16#3537, 16#3538, 16#3539, 16#3541,
16#3542, 16#3543, 16#3544, 16#3545, 16#3546, 16#3630, 16#3631,
16#3632, 16#3633, 16#3634, 16#3635, 16#3636, 16#3637, 16#3638,
16#3639, 16#3641, 16#3642, 16#3643, 16#3644, 16#3645, 16#3646,
16#3730, 16#3731, 16#3732, 16#3733, 16#3734, 16#3735, 16#3736,
16#3737, 16#3738, 16#3739, 16#3741, 16#3742, 16#3743, 16#3744,
16#3745, 16#3746, 16#3830, 16#3831, 16#3832, 16#3833, 16#3834,
16#3835, 16#3836, 16#3837, 16#3838, 16#3839, 16#3841, 16#3842,
16#3843, 16#3844, 16#3845, 16#3846, 16#3930, 16#3931, 16#3932,
16#3933, 16#3934, 16#3935, 16#3936, 16#3937, 16#3938, 16#3939,
16#3941, 16#3942, 16#3943, 16#3944, 16#3945, 16#3946, 16#4130,
16#4131, 16#4132, 16#4133, 16#4134, 16#4135, 16#4136, 16#4137,
16#4138, 16#4139, 16#4141, 16#4142, 16#4143, 16#4144, 16#4145,
16#4146, 16#4230, 16#4231, 16#4232, 16#4233, 16#4234, 16#4235,
16#4236, 16#4237, 16#4238, 16#4239, 16#4241, 16#4242, 16#4243,
16#4244, 16#4245, 16#4246, 16#4330, 16#4331, 16#4332, 16#4333,
16#4334, 16#4335, 16#4336, 16#4337, 16#4338, 16#4339, 16#4341,
16#4342, 16#4343, 16#4344, 16#4345, 16#4346, 16#4430, 16#4431,
16#4432, 16#4433, 16#4434, 16#4435, 16#4436, 16#4437, 16#4438,
16#4439, 16#4441, 16#4442, 16#4443, 16#4444, 16#4445, 16#4446,
16#4530, 16#4531, 16#4532, 16#4533, 16#4534, 16#4535, 16#4536,
16#4537, 16#4538, 16#4539, 16#4541, 16#4542, 16#4543, 16#4544,
16#4545, 16#4546, 16#4630, 16#4631, 16#4632, 16#4633, 16#4634,
16#4635, 16#4636, 16#4637, 16#4638, 16#4639, 16#4641, 16#4642,
16#4643, 16#4644, 16#4645, 16#4646).
在我的笔记本 i5 CPU M 520 @ 2.40GHz 在 10MB 块上进行测试时，它的性能为 90MB/s。但是优化在那里被发挥到了极致。如果使用 16 位查找，它也可以达到 97MB，但在这里发布太疯狂而且太长了。
【讨论】：
无需循环：Bits = bit_size(Bin), << Int:Bits >> = Bin, io:format("~.16B\n", [Int])【参考方案2】：
你可以这样做： [ hd(erlang:integer_to_list(Nibble, 16)) || >
这将返回一个包含二进制十六进制数字的列表（字符串）。虽然我怀疑此操作的效率是否会对系统的运行时产生任何影响，但您也可以让这个bin_to_hex 函数返回一个更易于构造的iolist，并且无论如何在输出时都会被展平。以下函数返回一个带有您提供的格式示例的 iolist：
bin_to_hex(Bin) when is_binary(Bin) ->
JoinableLength = byte_size(Bin) - 1,
<< Bytes:JoinableLength/binary, LastNibble1:4, LastNibble2:4 >> = Bin,
[ "<< ",
[ [ erlang:integer_to_list(Nibble1, 16), erlang:integer_to_list(Nibble2, 16), ", " ]
|| << Nibble1:4, Nibble2:4 >> <= Bytes ],
erlang:integer_to_list(LastNibble1, 16),
erlang:integer_to_list(LastNibble2, 16),
" >>" ].
它有点难看，但是会运行一次二进制文件并且不会遍历输出列表（否则我会使用 string:join 来获取散布的 ", " 序列）。如果这个函数不是某个进程的内部循环（我很难相信这个函数会成为你的瓶颈），那么你可能应该使用一些效率较低但更明显的代码，例如：
bin_to_hex(Bin) when is_binary(Bin) ->
"<< " ++ string:join([byte_to_hex(B) || << B >> <= Bin ],", ") ++ " >>".
byte_to_hex(<< N1:4, N2:4 >>) ->
[erlang:integer_to_list(N1, 16), erlang:integer_to_list(N2, 16)].
【讨论】：
【参考方案3】：bin_to_hex_list(Bin) when is_binary(Bin) ->
lists:flatten([integer_to_list(X,16) || <<X>> <= Bin]).
【讨论】：
这不做任何零填充，所以结果不正确。如果一个字节的前 4 位为 0，那么当需要两位时，它只会为该字节生成一位。【参考方案4】：改进@hairyhum
这会处理零填充
<< <<Y>> ||<<X:4>> <= Id, Y <- integer_to_list(X,16)>>
逆向变换
<<<<Z>> || <<X:8,Y:8>> <= Id,Z <- [binary_to_integer(<<X,Y>>,16)]>>, %%hex to binary
【讨论】：
【参考方案5】：这已经有一段时间没有看到任何行动了，但所有先前的解决方案似乎都过于复杂。 对我来说，这似乎要简单得多：
[begin if N < 10 -> 48 + N; true -> 87 + N end end || <<N:4>> <= Bin]
如果你喜欢它扩展一点：
[begin
if
N < 10 ->
48 + N; % 48 = $0
true ->
87 + N % 87 = ($a - 10)
end
end || <<N:4>> <= Bin]
【讨论】：
外层begin...end的目的是什么？【参考方案6】：这是我使用的另一个简短而快速的版本：
hexlify(Bin) when is_binary(Bin) ->
<< <<(hex(H)),(hex(L))>> || <<H:4,L:4>> <= Bin >>.
hex(C) when C < 10 -> $0 + C;
hex(C) -> $a + C - 10.
【讨论】：
【参考方案7】：如果您更喜欢制作二进制字符串而不是 erlang 默认列表字符串，则可以使用二进制理解语法，就像我在 sha1 生成代码中所做的那样：
1> << << if N >= 10 -> N -10 + $a;
1> true -> N + $0 end >>
1> || <<N:4>> <= crypto:hash(sha, "hello world") >>.
<<"2aae6c35c94fcfb415dbe95f408b9ce91ee846ed">>
与 python binascii.b2a_hex 相同：
>>> binascii.b2a_hex(sha.new('hello world').digest())
'2aae6c35c94fcfb415dbe95f408b9ce91ee846ed'
【讨论】：
【参考方案8】：从 OTP24 开始，有
1> binary:encode_hex(<<1,2,3,4,5,6,255>>).
<<"010203040506FF">>
【讨论】：
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