挥发性和序列点

Question

给定以下代码：

unsigned int global_flag = 0;

void exception_handle()
{
    global_flag = 1;
}

void func()
{
    /* access will cause exception which will assign global_flag = 1
       then execution continues */
    volatile unsigned int x = *(unsigned int *)(0x60000000U); /* memory protection unit configured to raise exception upon accessing this address */

    if (global_flag == 1)
    {
        /* some code */
    }
}

鉴于 volatile must not be reordered across sequence points:

The minimum requirement is that at a sequence point all previous accesses to volatile objects have stabilized and no subsequent accesses have occurred

并给出以下关于 sequence points 的信息：

sequence points occur in the following places ... (1) .. (2) .. (3) At the end of a full expression. This category includes expression statements (such as the assignment a=b;), return statements, the controlling expressions of if, switch, while, or do-while statements, and all three expressions in a for statement.

是否承诺volatile unsigned int x = *(unsigned int *)(0x60000000U);会在if (global_flag == 1)之前发生（在二进制asm中，CPU乱序执行与这里无关）？

根据上面的引文，volatile unsigned int x = *(unsigned int *)(0x60000000U);必须在下一个序列点结束之前计算，而volatile unsigned int x = *(unsigned int *)(0x60000000U);本身就是一个序列点，所以这意味着每个volatile 赋值是在赋值时求值的？

如果上述问题的答案是否定的，那么下一个序列点在if的end，是否意味着可以执行类似的事情:

if (global_flag == 1)
{
    volatile unsigned int x = *(unsigned int *)(0x60000000U);
    /* some code */
}

系统是嵌入式one-ARM cortex m0，单核，单线程应用。

Answer 1

Is it promised that volatile unsigned int x = *(unsigned int *)(ILLEGAL_ADDRESS); will take place before if (global_flag == 1)

来自信息性 C11 AnnexC（添加 newlines/formatting 以提高可读性）：

The following are the sequence points described in 5.1.2.3:
...
- Between the evaluation of a full expression and the next full expression to be evaluated.
- The following are full expressions:
- an initializer that is not part of a compound literal (6.7.9);
- the expression in an expression statement (6.8.3);
- the controlling expression of a selection statement (if or switch) (6.8.4);
- the controlling expression of a while or do statement (6.8.5);
- each of the (optional) expressions of a for statement (6.8.5.3);
- the (optional) expression in a return statement (6.8.6.4).

因为 *(unsigned int *)(ILLEGAL_ADDRESS); 是一个初始化器（赋值表达式）并且初始化器不是复合文字的一部分，所以它是一个完整的表达式。下一个完整表达式是 if 中的控制语句，因此在 if 和 x 的初始化之间有一个序列点。

来自著名的C11 5.1.2.3p6：

The least requirements on a conforming implementation are:

Accesses to volatile objects are evaluated strictly according to the rules of the abstract machine.
...

由于x是一个volatile对象，它被严格初始化为抽象机，所以在序列点之后它的右值必须等于*(unsigned int *)(ILLEGAL_ADDRESS)操作的结果。

所以是的，x 对象的初始化必须发生在 if.

中的控制表达式之前

关于未定义的行为，引用自C11 6.5.3.2p4：

If an invalid value has been assigned to the pointer, the behavior of the unary * operator is undefined.

正如您评论的那样：

accessing address 0x60000000 is not permitted in my system memory model

可以推断出 (unsigned int*)0x60000000 是无效指针，因此一元 * 运算符应该 spawn dragons.

Answer 2

在您的代码片段中，变量 global_flag 不是易变的，因此没有什么可以阻止编译器跨序列点移动对 global_flag 的访问，或者如果情况允许，则完全删除它。讨论访问x和访问global_flag的顺序是没有意义的，因为后者不是可观察事件，只有前者是。

（另请注意，表达式 *(unsigned int *)(0x60000000U) 中没有 volatile 限定符。我认为这确实是您希望特别对待的表达式，但您的代码不会那样做。允许编译器生成预先评估 *(unsigned int *)(0x60000000U) 的代码，然后执行其盘子上的大量其他内容，然后将获得的值分配给 x，这将满足以下约束C 标准放在 volatile 左值上。）

如果您的代码段有 unsigned int volatile global_flag = 0; 和 *(volatile unsigned int *)(0x60000000U)，那么“是否承诺……”这个问题的答案将是明确的“是”。

挥发性和序列点

Volatile and sequence point

c

volatile

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