[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(16) 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。