正在更新双操作原子 [英] Is Updating double operation atomic
问题描述
在Java中，更新double和long变量可能不是原子的，因为double / long被视为两个单独的32位变量。
In Java, updating double and long variable may not be atomic, as double/long are being treated as two separate 32 bits variables.
http://java.sun.com/docs/books/jls/second_edition/html /memory.doc.html#28733
在C ++中，如果我使用32位Intel Processor + Microsoft Visual C ++编译器， ）操作原子？
In C++, if I am using 32 bit Intel Processor + Microsoft Visual C++ compiler, is updating double (8 byte) operation atomic?
我不能在这个行为上找到太多的规范。
I cannot find much specification mention on this behavior.
当我说原子变量我的意思是：
When I say "atomic variable", here is what I mean :
线程A试图写1到变量x。
线程B试图写2到变量x。
Thread A trying to write 1 to variable x. Thread B trying to write 2 to variable x.
我们将从变量x得到值1或2，但不是未定义的值。
We shall get value 1 or 2 out from variable x, but not an undefined value.
推荐答案
这是硬件特定的，取决于架构。对于x86和x86_64，如果它们对齐，则8个字节的写或读保证是原子的。引用英特尔架构内存订购白皮书：
This is hardware specific and depends an the architecture. For x86 and x86_64 8 byte writes or reads are guaranteed to be atomic, if they are aligned. Quoting from the Intel Architecture Memory Ordering White Paper:
英特尔64内存排序保证
每个以下
存储器访问指令，
组成存储器操作出现
作为单个存储器访问
执行，而不考虑存储器类型：
Intel 64 memory ordering guarantees that for each of the following memory-access instructions, the constituent memory operation appears to execute as a single memory access regardless of memory type:
读取或写入单个字节的指令。
Instructions that read or write a single byte.
读取或写入字（2字节）
Instructions that read or write a word (2 bytes) whose address is aligned on a 2 byte boundary.
读取或写入地址为
的双字（4字节）的指令
Instructions that read or write a doubleword (4 bytes) whose address is aligned on a 4 byte boundary.
所有锁定的指令（隐式
锁定xchg指令和其他
读取 - 修改 - 写入指令与
锁定前缀）是不可分割的和
不间断的加载序列
后跟存储，而不管
存储器类型和对齐。
All locked instructions (the implicitly locked xchg instruction and other read-modify-write instructions with a lock prefix) are an indivisible and uninterruptible sequence of load(s) followed by store(s) regardless of memory type and alignment.
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