[C++关键字] alignof & alignas 内存对齐 sizeof 占内存大小

直接上代码测试是入门神器，以结构体为例，解释“对齐”和“补齐”概念。

#include <iostream>
struct Empty {};
struct Foo {
    int f2;
    double f1;
    char c;
};
struct alignas() FooNew
{
    int f2;
    double f1;
    char c;
};

int main()
{
    std::cout << "alignment of empty class: " << alignof(Empty) << '\n'
    << "alignment of pointer : "    << alignof(int*)  << '\n'
    << "alignment of char : "       << alignof(char)  << '\n'
    << "alignment of Foo : "        << alignof(Foo)   << '\n'
    << "Size of Foo: " << sizeof(Foo) << '\n'
    << "alignment of FooNew : "        << alignof(FooNew)   << '\n';
}

输出结果是：

alignment of empty class: 1
alignment of pointer : 8
alignment of char : 1
alignment of Foo : 8
Size of Foo: 24
alignment of FooNew : 16

总之，对齐是某种类型的初始位置在内存上的限定，补齐是对该类型大小的限定，两者共同组成了该类型在内存上的排布规则，提高操作效率。