这个声明如何调用Most Vexing Parse？ [英] How does this declaration invoke the Most Vexing Parse?

问题描述

请考虑以下程序：

#include <fstream>

struct A {};

int main(int argc, char** argv) {
    A a(std::fstream(argv[1]));
}

在C ++ 1y模式中的Clang假定MVP被调用， a href =http://coliru.stacked-crooked.com/a/393a356e98041cf2> a 被解析为函数声明：

Clang in C++1y mode reckons that the MVP is invoked such that a is parsed as a function declaration:

clang++ -std=c++1y -O3 -Wall -Wextra -pedantic-errors -pthread main.cpp && ./a.out

main.cpp:6:8: warning: parentheses were disambiguated as a function declaration [-Wvexing-parse]

    A a(std::fstream(argv[1]));

       ^~~~~~~~~~~~~~~~~~~~~~~

main.cpp:6:9: note: add a pair of parentheses to declare a variable

    A a(std::fstream(argv[1]));

        ^

        (                    )

MVP，但不是在这种情况下： argv [1] 显然是一个表达式，并且没有类型之前，所以这行如何解析为一个函数声明？

I understand the MVP, but not in this instance: argv[1] is clearly an expression, and there's no type before it, so how could this line be parsed as a function declaration?

是 argv [1] 的语义解释，它会将行解释为对象声明，直到编译器已经选择将该行解析为函数声明之后才会发生？或者是一个Clang bug？或者是通过对标记 argv [1] 的一些解释，我错过了吗？

Is it that the semantic interpretation on argv[1], that would disambiguate the line as an object declaration, doesn't occur until after the compiler has already chosen to parse the line as a function declaration? Or is it a Clang bug? Or is it perfectly normal through some interpretation of the tokens argv [ 1 ] that I'm missing?

推荐答案

我认为它被解析为

A a(std::fstream argv[1]);

一个函数接受1 std :: fstream 的数组，其中额外的括号是多余的。

i.e. a function that takes in an array of 1 std::fstream, where the extra parentheses are redundant.

当然，在现实中，数组参数衰减到指针，所以你最终以语义是：

Of course, in reality, that array parameter decays to a pointer, so what you end up with semantically is:

A a(std::fstream* argv);

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