std::atomic\u bool值的严格转换

我有一个非常简单的逻辑算术代码,其中包含从 std::atomic_bool .

#include <iostream>
#include <atomic>

int main() {
    uint16_t v1 = 0x1122;
    uint16_t v2 = 0xaaff;
    std::atomic_bool flag1(false);

    uint16_t r1 = v1 | v2;
    std::cout << std::hex << r1 << std::endl;

    uint16_t r2 = static_cast<uint16_t>(flag1.load()) | static_cast<uint16_t>(0xaaff);
    std::cout << std::hex << r2 << std::endl;

    std::cout << __VERSION__ << std::endl;
}

代码示例为 here .编译行: g++ -std=c++17 -O3 -Wall -pedantic -Wconversion -pthread main.cpp && ./a.out .

根据STD API, load() 应返回存储在原子中的带下划线的类型。所以 flag1.load() 应该返回 bool 。但是,编译器会发出警告,要求它转换 int 到 uint16_t :

main.cpp:13:55: warning: conversion from 'int' to 'uint16_t' {aka 'short unsigned int'} may change value [-Wconversion]
     uint16_t r2 = static_cast<uint16_t>(flag1.load()) | static_cast<uint16_t>(0xaaff);
                   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

它究竟在哪里进行这种转换?两岸 | 已转换为 uint16\u t 。为什么它仍在打印警告?