是否可以进行此编译器转换? [英] Is this compiler transformation allowed?

问题描述

考虑以下代码，其中x和y是整数:

Consider this code, where x and y are integers:

if (x)
    y = 42;

是否允许以下​​编译器转换?

Is the following compiler transformation allowed ?

int tmp = y;
y = 42;

if (!x)
    y = tmp;

上下文:

这是来自Bjarne Stroustrup的常见问题解答:

This is from Bjarne Stroustrup's FAQ:

// start with x==0 and y==0

if (x) y = 1;   // Thread 1 

if (y) x = 1;   // Thread 2

常见问题解答指出这是免费的数据竞赛；如果x和y均为0，则不应将任何var写入.
但是，如果允许转换怎么办?

The FAQ states this is data race free; with x and y both 0, none of the vars should be written to.
But what if the transformation is allowed ?

推荐答案

与我在不正确的注释中写的不同，如果y可能在线程之间共享，并且编译器无法证明线程中存在任何现有的UB，则实际上不允许这种转换.原始代码.

Unlike I wrote in my incorrect comment, this transformation is actually not allowed if y is potentially shared between threads and the compiler cannot prove any existing UB in the original code.

该标准明确规定:

编译器转换，将分配引入到潜在的共享内存位置，该位置 不会被抽象机器修改的对象通常会被该标准排除，因为这样的 在抽象机器的情况下，分配可能会通过不同的线程覆盖另一个分配 执行过程不会遇到数据争用.

Compiler transformations that introduce assignments to a potentially shared memory location that would not be modified by the abstract machine are generally precluded by this standard, since such an assignment might overwrite another assignment by a different thread in cases in which an abstract machine execution would not have encountered a data race.

N3337中的[intro.multithread](1.10/22)，N4141中的(1.10/25).

[intro.multithread] (1.10/22) in N3337, (1.10/25) in N4141.

因此，如果x始终为0，则原始代码将是无竞争的，而转换后的代码则不会.因此，这种转换是不合法的.

So if x is always 0, the original code would be race-free, while the transformed one wouldn't. Thus the transformation is not legal.

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