C11 - 编译器忽略潜在的无限循环

Question

假设以下代码

struct a {
    unsigned cntr;
};

void boo(struct a *v) {
    v->cntr++;
    while(v->cntr > 1);
}

由于 C11 标准中的以下语句，我想知道是否允许编译器省略 boo() 内的 while 循环:

An iteration statement whose controlling expression is not a constant expression,¹⁵⁶⁾ that performs no input/output operations, does not access volatile objects, and performs no synchronization or atomic operations in its body, controlling expression, or (in the case of a for statement) its expression-3, may be assumed by the implementation to terminate.¹⁵⁷⁾

---

¹⁵⁷⁾_{This is intended to allow compiler transformations such as removal of empty loops even when termination cannot be proven.}

控制表达式中的v->cntr可以被视为同步，因为v可能是指向全局结构的指针可以从外部修改(例如通过另一个线程)？

附加问题。如果v未定义为 volatile ，编译器是否允许在每次迭代时不重新读取v->cntr？

Answer 1

Can v->cntr, in the controlling expression, be considered as a synchronization

没有。

来自https://port70.net/~nsz/c/c11/n1570.html#5.1.2.4p5 :

The library defines a number of atomic operations (7.17) and operations on mutexes (7.26.4) that are specially identified as synchronization operations.

基本上，来自 stdatomic.h 的函数和来自 thread.h 的 mtx_* 都是同步操作。

since v may be a pointer to a global structure which can be modified externally (for example by another thread)?

没关系。对我来说，这些假设听起来像是不允许许多合理的优化，我不希望我的编译器做出这样的假设。

如果 v 在另一个线程中被修改，那么它将是无序的，这只会导致未定义的行为 https://port70.net/~nsz/c/c11/n1570.html#5.1.2.4p25 .

Is the compiler allowed not to re-read v->cntr on each iteration if v is not defined as volatile?

是的。