什么是begin…endinErlang用于什么?-Whatisbegin…endinErlangusedfor?

5  

Macros, for example:

宏,例如:

-define(M(A, B),
    begin
        C = foo(),
        bar(A, B, C)
    end).

6  

To evaluate a catch (always the same idea to have multiple expression reduced to one)

为了评估一个捕获(总是相同的想法，将多个表达式简化为一个)

Res = (catch
    begin
        C = foo(Bar),
        io:format("evaluation of C ok~n"),
        D = bar(A, B, C)
     end),

3  

As previous answerers mentioned, this construct is used whenever you need to have multiple expressions but only one is allowed.

正如前面提到的答案一样，无论何时需要多个表达式，但只允许一个表达式，都要使用这个结构。

However, the majority of such cases would be considered a stinky style. I can remember only a few of places where a single expression is expected: an argument in a function call, catch expression, case of, try of and list comprehension. All of them except for list comprehension shouldn't be used with begin end construct because the variables are leaking to the outer scope probably causing the subsequent bindings to become matches.

然而，大多数此类案件被认为是一种臭气熏天的风格。我只记得有几个地方需要一个表达式:函数调用中的一个参数、catch表达式、大小写、try of和list comprehension。除了列表理解之外，它们都不应该与begin end结构一起使用，因为变量泄漏到外部范围，可能导致后续绑定成为匹配。

List comprehension expression is different because it is transformed to a separate function with its own scope and no variable introduced in begin end leaks to the outer scope.

列表理解表达式是不同的，因为它被转换为一个单独的函数，并且在开始时没有引入任何变量，从而在外部作用域中引入。

0  

According to erlang documentation is it block expression that evaluates each expression but returns only the last one.

根据erlang文档，它的块表达式计算每个表达式，但只返回最后一个表达式。

See this example (not using block expression):

参见这个示例(不使用块表达式):

A = 1,
case A + 1 of
    3 ->
        ok;
    _->
        nop
end.

% returns ok

Now you can define A within the case argument using block expression:

现在您可以使用块表达式在case参数中定义A:

case begin A = 1, A + 1 end of
    3 ->
        ok;
    _->
        nop
end.

%returns ok

That evaluates A = 1, then returns the result of A + 1.

它计算A = 1，然后返回A + 1的结果。

Now we know that this will not work:

现在我们知道这行不通:

case A = 1, A + 1 of
    3 ->
        ok;
    _->
        nop
end.

% returns syntax error before: ','