std::assume_aligned

 

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定义在头文件 <memory> 中
template< std::size_t N, class T > constexpr T* assume_aligned( T* ptr );		(自 C++20 起)

告知实现对象 ptr 指向的至少对齐到 N。实现可以使用此信息生成更高效的代码，但它可能只会在通过 assume_aligned 的返回值访问对象时做出此假设。

N 必须是 2 的幂。如果 ptr 不指向类型为 T 的对象（忽略每个级别的 cv 限定），或者对象的对齐方式不至少为 N，则行为未定义。

返回值

ptr.

异常

不抛出任何内容。

注释

为了确保程序从 assume_aligned 启用的优化中受益，通过其返回值访问对象非常重要

void f(int* p)
{
    int* p1 = std::assume_aligned<256>(p);
    // Use p1, not p, to ensure benefit from the alignment assumption.
    // However, the program has undefined behavior if p is not aligned
    // regardless of whether p1 is used.
}

程序有责任确保对齐假设实际上成立。调用 assume_aligned 不会导致编译器验证或强制执行此操作。

功能测试 宏	值	Std	功能
__cpp_lib_assume_aligned	201811L	(C++20)	std::assume_aligned

另请参见

alignof 运算符(C++11)	查询类型的对齐要求
alignas 说明符(C++11)	指定变量的存储空间应按特定数量对齐
aligned_storage (C++11)(C++23 中已弃用)	定义适合用作给定大小类型的未初始化存储空间的类型 (类模板)
align (C++11)	在缓冲区中对齐指针 (函数)
[[assume(expression)]](C++23)	指定在给定点，表达式将始终计算为true。 (属性说明符)