volatile const int x = 42;const volatile int *p = &x;*(volatile int *)p = 8; // Does this line introduce undefined behavior?*(int *)p = 16; // And what about this one?
Compilable code
解决方法 当您尝试修改“初始”const对象时,它是未定义的行为(对于两个语句).从C11(N1570)6.7.3 / p6类型限定符(强调我的):If an attempt is made to modify an object defined with a
const-qualifIEd type through use of an lvalue with non-const-qualifIEd
type,the behavior is undefined.
为了完整性,它可能值得添加,标准也说:
If an attempt is made to refer to an object defined with a
volatile-qualifIEd type through use of an lvalue with
non-volatile-qualifIEd type,the behavior is undefined.
因此后面的陈述,即:
*(int *)p = 16;
对于第二个短语也是未定义的(它是“双UB”).
我相信C的规则是相同的,但不要拥有C 14的副本来确认.
总结以上是内存溢出为你收集整理的c – 可以将volatile volatile的间接变化视为未定义的行为吗?全部内容,希望文章能够帮你解决c – 可以将volatile volatile的间接变化视为未定义的行为吗?所遇到的程序开发问题。
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