問題描述

我正在學習 C++11，我偶然發現了統一初始化程序.

I'm studying C++11 and I stumbled upon uniform initializers.

我不明白以下應該顯示最令人煩惱的解析"歧義的代碼:

I don't understand the following code which should show the "most vexing parse" ambiguity:

#include<iostream>


class Timer
{
public:
  Timer() {}
};

int main() 
{

  auto dv = Timer(); // What is Timer() ? And what type is dv?

  int time_keeper(Timer()); // This is a function right? And why isn't the argument " Timer (*) ()" ?



  return 0;
}

推薦答案

這里:

auto dv = Timer();

您有一個名為 dv 的 Timer 類型的對象，它正在從臨時對象(= 右側的表達式)復制初始化代碼> 符號).

You have an object of type Timer called dv that is being copy-initialized from a temporary (the expression on the right side of the = sign).

當使用 auto 聲明一個變量時，該變量的類型與初始化它的表達式的類型相同——這里不考慮 cv 限定符和引用.

When using auto to declare a variable, the type of that variable is the same as the type of the expression that initializes it - not considering cv-qualifiers and references here.

在您的情況下，初始化 dv 的表達式的類型為 Timer，因此 dv 的類型為 Timer.

In your case, the expression that initializes dv has type Timer, and so dv has type Timer.

這里:

int time_keeper(Timer());

您聲明了一個名為 time_keeper 的函數，它返回一個 int 并將指針作為它的輸入，該函數返回一個 定時器，不帶參數.

You declare a function called time_keeper that returns an int and takes as its input a pointer to a function which returns a Timer and takes no argument.

為什么不是參數 Timer (*) () ?

當作為參數傳遞時，函數衰減為指針，所以time_keeper的類型實際上是int(Timer(*)()).

Functions decay to pointers when passed as an argument, so the type of time_keeper is actually int(Timer(*)()).

為了說服自己，你可以嘗試編譯這個小程序:

To convince yourself, you could try compiling this little program:

#include <type_traits>

struct Timer { };
int main()
{
    int time_keeper(Timer());
    static_assert(
        std::is_same<
            decltype(time_keeper), 
            int(Timer(*)())
        >::value, 
        "This should not fire!");
}

這里有一個referar">noliver>>.

Here is a live example.

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