自然对齐+ volatile = C ++ 11中的原子？ [英] Natural alignment + volatile = atomic in C++11?

问题描述

1）以下是自然对齐指针的声明：

1) Is the following declaration of a naturally aligned pointer:

alignas(sizeof(void *)) volatile void * p;

相当于

std::atomic<void *> 

在C ++ 11中？

2）更准确地说，假设这种类型的指针将以与C ++ 11中的std :: atomic相同的方式工作是否正确？

2) Saying more exactly, is it correct to assume that this type of pointer will work in the same way as std::atomic in C++11?

推荐答案

不，volatile不保证该位置将被原子地读写，只是编译器无法优化多个读写。

No, volatile does not guarantee that the location will be written or read atomically, just the the compiler can't optimise out multiple reads and writes.

在某些体系结构上，如果对齐正确，处理器将自动进行读写操作，但这不是通用的，甚至不能通过一系列处理器来保证。在可能的情况下，原子的内部实现将利用体系结构功能和原子指令修饰符，因此，如果您是指原子，为什么 不 使用原子？

On certain architectures, the processor will atomically read or write if aligned correctly, but that is not universal or even guaranteed through a family of processors. Where it can, the internal implementation of atomic will take advantage of architectural features and atomic instruction modifiers, so why not use atomic, if you mean atomic?

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