为什么将括号与默认构造函数一起使用会导致创建变量？

Question

看完 Louis Brandy 在 CppCon 2017 上的演讲后，我震惊地发现这段代码实际上可以编译：

#include <string>

int main() {

    std::string(foo);

    return 0;
}

并且出于某种原因 std::string(foo) 它与 std::string foo 相同，即声明一个变量。我发现它绝对违反直觉，并且看不出 C++ 以这种方式工作的任何理由。我希望这会给出有关未定义标识符 foo.

的错误

它实际上使像 token1(token2) 这样的表达式比我之前想象的有更多可能的解释。

所以我的问题是：这种恐怖的原因是什么？什么时候真正需要这条规则？

P.S。抱歉，标题措辞不佳，请随时更改！

Answer 1

因为这个问题被标记了 language-lawyer, the direct answer is that, from [stmt.ambig]:

There is an ambiguity in the grammar involving expression-statements and declarations: An expression-statement with a function-style explicit type conversion as its leftmost subexpression can be indistinguishable from a declaration where the first declarator starts with a (. In those cases the statement is a declaration.

同样，对于函数，在 [dcl.ambig.res]:

The ambiguity arising from the similarity between a function-style cast and a declaration mentioned in [stmt.ambig] can also occur in the context of a declaration. In that context, the choice is between a function declaration with a redundant set of parentheses around a parameter name and an object declaration with a function-style cast as the initializer. Just as for the ambiguities mentioned in [stmt.ambig], the resolution is to consider any construct that could possibly be a declaration a declaration.

因此：

Why oh why is std::string("foo") so different from std::string(foo)

前者不能声明。后者可以是一个声明，带有一组多余的括号。因此，前者不是声明，而后者是 .

潜在的问题是，在语法上，声明符可以以 ( 开头，这可能使其与函数式显式类型转换无法区分。与其想出任意复杂的规则来尝试确定用户的意思，语言只是选择一个，并且用户很容易修复代码以实际执行他的意思。

为什么将括号与默认构造函数一起使用会导致创建变量？

Why does using parentheses with a default constructor result in creation of the variable?

c++

most-vexing-parse

language-lawyer