小编Mat*_*att的帖子

这是一个最小的设置，有 2 个表a，b每个表有 3 行：

CREATE TABLE a (
    id SERIAL PRIMARY KEY,
    value TEXT
);
CREATE INDEX ON a (value);

CREATE TABLE b (
    id SERIAL PRIMARY KEY,
    value TEXT
);
CREATE INDEX ON b (value);

INSERT INTO a (value) VALUES ('x'), ('y'),        (NULL);
INSERT INTO b (value) VALUES        ('y'), ('z'), (NULL);

这是一个按预期工作正常的 LEFT JOIN：

SELECT * FROM a
LEFT JOIN b ON a.value IS NOT DISTINCT FROM b.value;

带输出：

 id | value | id …

以下程序编译时不会出现警告-O0：

#include <iostream>\n\nstruct Foo\n{\n  int const& x_;\n  inline operator bool() const { return true; }\n  Foo(int const& x):x_{x} { }\n  Foo(Foo const&) = delete;\n  Foo& operator=(Foo const&) = delete;\n};\n\nint main()\n{\n  if (Foo const& foo = Foo(3))\n    std::cout << foo.x_ << std::endl;\n\n  return 0;\n}\n

但是，使用-O1或更高版本会发出警告：

maybe-uninitialized.cpp: In function \xe2\x80\x98int main()\xe2\x80\x99:\nmaybe-uninitialized.cpp:15:22: warning: \xe2\x80\x98<anonymous>\xe2\x80\x99 is used uninitialized in this function [-Wuninitialized]\n std::cout << foo.x_ << std::endl;\n

如何摆脱这个警告-O1？

这样做的动机是CHECK(x)宏必须捕获 const 引用而不是值，以免触发析构函数、复制构造函数等以及打印出值。 …

我有5个数组:

["A", "B", "C"]
["A", "B", "C", "D", "E"]
["A"]
["A", "B", "C", "D", "E", "F"]
["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O"]

我想创建一个每个组合的列表:

["AAAAA","AAAAB","AAAAC", "AAAAD"...
 "BAAAA","BAAAB","BAAAC", "BAAAD"...]

static constexpr size_t Foo<>::sum()在C++ 11中编写下面的方法最简洁的方法是什么？这适用于C++ 17编译器,但我正在寻找一种适用于C++ 11模式的g ++,clang和Visual Studio 2015的方法.

#include <iostream>
#include <type_traits>

template<typename T,size_t N>
class Foo
{
  public:
    static constexpr size_t sum();
};

template<typename>
struct is_foo : std::false_type { };
template<typename T,size_t N>
struct is_foo<Foo<T,N>> : std::true_type { };

template<typename T,size_t N>
constexpr size_t Foo<T,N>::sum()
{
    if constexpr (is_foo<T>::value)
        return N + T::sum();
    else
        return N;
}

int main()
{
    constexpr size_t sum = Foo<Foo<Foo<double,3>,4>,5>::sum(); // 12 = 3+4+5
    std::cout << "sum = " << sum …