警告：缩小转换 C++11

Question

g++ 4.9.0 -O2 -std=c++11

template<class T>
struct vec3 {
    T x, y, z;
    vec3() = default;
    vec3(const vec3<T> &other) = default;
    vec3(T xx, T yy, T zz) { x = xx; y = yy; z = zz; }
    vec3<T> operator-(const vec3<T> &other) { 
      return vec3<T>{ x - other.x, y - other.y, z - other.z }; 
    }
};

int main() {
    vec3<char> pos{ 0, 0, 0 };
    vec3<char> newPos{ 0, 0, 0 };
    auto p = pos - newPos;

    return 0;
}

我收到警告：

!!warning: narrowing conversion of ‘(((int)((vec3<char>*)this)->vec3<char>::x) - ((int)other.vec3<char>::x))’ from ‘int’ to ‘char’ inside { } [-Wnarrowing]

但是如果我在 operator- 函数中使用 (...) insted of {...} 来做到这一点，警告就会消失。为什么？

Answer 1

这里发生了几件事。首先，{...} 语法禁止隐式收缩转换。所以简单的解决方法是将大括号更改为圆括号：

vec3<T> operator-(const vec3<T> &other) { 
  return vec3<T>( x - other.x, y - other.y, z - other.z ); 
}

第二件事是，"eh? char minus a char is a char, what is the problem?!"这里的答案是C/C++想要使用自然大小进行算术运算。这就是您在错误消息中看到 (int) 强制转换的原因。 Here is a good explanation 为什么这样做（以防 Whosebug 的答案消失，他引用了 C11 标准的 6.3.1.1）。

因此，修复代码的另一种方法是：

vec3<T> operator-(const vec3<T> &other) { 
  return vec3<T>{
    static_cast<char>(x - other.x),
    static_cast<char>(y - other.y),
    static_cast<char>(z - other.z)
    };
}

顺便说一下，Effective Modern C++ 中的第 7 项让我相信，有时 () 更适合初始化，有时 {} 更好。有时你不得不耸耸肩，然后使用另一个。

Answer 2

首先，为什么缩小？来自§5/10:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

— [..]

— Otherwise, the integral promotions (4.5) shall be performed on both operands.

其中积分提升在 4.5/1 中定义：

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.

在我们的例子中，decltype(char + char) 是 int，因为 char 的转化排名低于 int，因此两者都被提升为 int在调用 operator+ 之前。现在，我们有 ints，我们正在将其传递给采用 chars 的构造函数。根据定义（§8.5.4/7，特别是 7.4）：

A narrowing conversion is an implicit conversion

(7.4) — from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

根据 §8.5.4/3 明确禁止列表初始化（强调我的，“见下文”实际上指的是我刚刚在上面复制的内容）：

List-initialization of an object or reference of type T is defined as follows

— [..]

— Otherwise, if T is a class type, constructors are considered. The applicable constructors are enumerated and the best one is chosen through overload resolution (13.3, 13.3.1.7). If a narrowing conversion (see below) is required to convert any of the arguments, the program is ill-formed. [...]

这就是您的 vec3<T>{int, int, int} 给您警告的原因：由于整数提升需要对所有表达式进行收缩转换，程序格式错误。现在，关于“格式错误”的声明只出现在列表初始化的上下文中。这就是为什么如果您在没有 {}s 的情况下初始化向量，您将看不到该警告：

vec3<T> operator-(const vec3<T> &other) { 
    // totally OK: implicit conversion from int --> char is allowed here
    return vec3<T>( x - other.x, y - other.y, z - other.z );
}

至于解决这个问题 - 只调用构造函数而不进行列表初始化可能是最简单的解决方案。或者，您可以继续使用列表初始化并仅对构造函数进行模板化：

template <typename A, typename B, typename C>
vec3(A xx, B yy, C zz)
: x(xx) // note these all have to be ()s and not {}s for the same reason
, y(yy)
, z(yy)
{ }

警告：缩小转换 C++11

warning: narrowing conversion C++11

c++

templates

narrowing

c++11