Erlang：变量未绑定 [英] Erlang: variable is unbound

问题描述

以下为什么说变量不受限制？

Why is the following saying variable unbound?

9> {<<A:Length/binary, Rest/binary>>, Length} = {<<1,2,3,4,5>>, 3}.     
* 1: variable 'Length' is unbound

很明显，长度应该为3。

我正在尝试使用类似模式匹配的函数，即：

I am trying to have a function with similar pattern matching, ie.:

parse(<<Body:Length/binary, Rest/binary>>, Length) ->

但是如果由于相同的原因失败。如何实现我想要的模式匹配？

But if fails with the same reason. How can I achieve the pattern matching I want?

我真正想要实现的是解析传入的TCP流数据包

What I am really trying to achieve is parse in incoming tcp stream packets as LTV(Length, Type, Value).

在我解析了长度和类型之后的某个时候，我只想准备最多 Length 个字节作为值，其余部分可能用于下一个LTV。

At some point after I parse the the Length and the Type, I want to ready only up to Length number of bytes as the value, as the rest will probably be for the next LTV.

所以我的 parse_value 函数是这样的：

So my parse_value function is like this:

parse_value(Value0, Left, Callback = {Module, Function},
       {length, Length, type, Type, value, Value1}) when byte_size(Value0) >= Left ->
    <<Value2:Left/binary, Rest/binary>> = Value0,
    Module:Function({length, Length, type, Type, value, lists:reverse([Value2 | Value1])}),
    if 
    Rest =:= <<>> ->
        {?MODULE, parse, {}};
    true ->
        parse(Rest, Callback, {})
    end;
parse_value(Value0, Left, _, {length, Length, type, Type, value, Value1}) ->
    {?MODULE, parse_value, Left - byte_size(Value0), {length, Length, type, Type, value, [Value0 | Value1]}}.

如果可以进行模式匹配，可以将其分解为更令人愉悦的外观。

If I could do the pattern matching, I could break it up to something more pleasant to the eye.

推荐答案

模式匹配的规则是，如果变量X出现在两个子模式中，例如{X，X}，或者{X，[X]}或类似的字符，则它们在两个位置必须具有相同的值，但每个子模式的匹配仍在相同的输入环境中完成-一侧的绑定不会转移到另一侧。相等性检查将在概念上完成，就像您在{X，X2}上进行了匹配并添加了保护X =：= X2一样。这意味着即使没有将元组中的Length字段设为最左侧的元素，也不能将其用作二进制模式的输入。

The rules for pattern matching are that if a variable X occurs in two subpatterns, as in {X, X}, or {X, [X]}, or similar, then they have to have the same value in both positions, but the matching of each subpattern is still done in the same input environment - bindings from one side do not carry over to the other. The equality check is conceptually done afterwards, as if you had matched on {X, X2} and added a guard X =:= X2. This means that your Length field in the tuple cannot be used as input to the binary pattern, not even if you make it the leftmost element.

但是，在是二进制模式，绑定在字段中的变量可以从左到右在其后的其他字段中使用。因此，以下工作（在二进制文件中使用前导32位大小的字段）：

However, within a binary pattern, variables bound in a field can be used in other fields following it, left-to-right. Therefore, the following works (using a leading 32-bit size field in the binary):

1> <<Length:32, A:Length/binary, Rest/binary>> = <<0,0,0,3,1,2,3,4,5>>.
<<0,0,0,3,1,2,3,4,5>>
2> A.
<<1,2,3>>
3> Rest.                                                               
<<4,5>>

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