为什么编译器不断警告我从“int”缩小到“uint8”

Why is compiler keep warning me that narrowed from `int` to `uint8`?

本文关键字：int uint8 缩小 编译器 警告 为什么      更新时间：2023-10-16

我再次检查了一遍，我确定我没有uint8投int 隐含在我的代码中，向前或向后

。 
// main.cpp
#include <iostream>
using std::cout, std::endl;
using uint8 = unsigned char;
struct Vector {
  uint8 x, y, z;
  Vector operator+(const Vector& v) const {
    return Vector{this->x + v.x, this->y + v.y, this->z + v.z};
  };
  void operator+=(const Vector&);
  void operator-(const Vector&) const;
  Vector operator-=(const Vector&);
  Vector operator*(const uint8 scaler) const;
  Vector Corss(const Vector&, const Vector&) const;
  void Corss(const Vector&);
  Vector Dot(const Vector&, const Vector&) const;
  void Dot(const Vector&);
};
std::ostream& operator<<(std::ostream& os, Vector& vec) {
  return os << "(" << (unsigned)vec.x << "," << (unsigned)vec.y << ","
            << (unsigned)vec.z << ")";
}
int main() {
  Vector v1{1, 2, 3};
  Vector v2{1, 1, 1};
  Vector v3 = v1 + v2;
  cout << v1 << endl;
  cout << v2 << endl;
  cout << v3 << endl;
}

提示：

$g++ main.cpp -std=c++17 -O0 -g -Wall -Wextra
main.cpp: In member function 'Vector Vector::operator+(const Vector&) const':
main.cpp:10:27: warning: narrowing conversion of '(((int)((const Vector*)this)->Vector::x) + ((int)v.Vector::x))' from 'int' to 'uint8 {aka unsigned char}' inside { } [-Wnarrowing]
     return Vector{this->x + v.x, this->y + v.y, this->z + v.z};
                   ~~~~~~~~^~~~~
main.cpp:10:42: warning: narrowing conversion of '(((int)((const Vector*)this)->Vector::y) + ((int)v.Vector::y))' from 'int' to 'uint8 {aka unsigned char}' inside { } [-Wnarrowing]
     return Vector{this->x + v.x, this->y + v.y, this->z + v.z};
                                  ~~~~~~~~^~~~~
main.cpp:10:57: warning: narrowing conversion of '(((int)((const Vector*)this)->Vector::z) + ((int)v.Vector::z))' from 'int' to 'uint8 {aka unsigned char}' inside { } [-Wnarrowing]
     return Vector{this->x + v.x, this->y + v.y, this->z + v.z};
                                                 ~~~~~~~~^~~~~
$./a.out
(1,2,3)
(1,1,1)
(2,3,4)

表达式this->x + v.x&c.中的每个项都会自动加宽为一个int，并且该表达式的类型是int。这是因为+的论据比int窄。然后，编译器会向您发出警告，因为您正在初始化类型比 int 窄Vector元素。

这是C++生活中的事实。你可以得到同样的效果，比如

auto a = 'a' + 'b';

a是一种int类型。

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