是否使用未初始化的变量 UB 的地址? [英] Is using the address of an uninitialized variable UB?

问题描述

这个小码是UB吗?

void Test()
{
  int bar;
  printf("%p", &bar);  
}

IMO 不是 UB，但我想要一些其他的意见.

IMO it's not UB, but I'd like some other opinions.

它只是打印bar的地址，即使bar从未被初始化.

It simply prints the address of bar, even if bar has never been initialized.

推荐答案

TL:DR 不，您的代码不会像您可能那样使用任何未初始化来调用 UB有想过.

TL:DR No, your code does not invoke UB by using anything uninitialized, as you might have thought.

a(ny) 变量的地址(在这种情况下是自动的)有一个 defined 值，所以无论变量本身是否被初始化，变量的地址都是一个 定义值.您可以利用该值.(如果您不处理指针和进行双重取消引用.:) )

The address of a(ny) variable (automatic, in this case) has a defined value, so irrespective of whether the variable itself is initialized or not, the address of the variable is a defined value. You can make use of that value. ( if you're not dealing with pointers and doing double-dereference. :) )

也就是说，严格来说，你应该写

That said, strictly speaking, you should write

 printf("%p", (void *)&bar);

as %p 需要一个指向 void 和 printf() 的类型指针的参数是一个可变参数函数，没有提升 (转换).否则，这是一个明确定义的行为.

as %p expects an argument of type pointer to void and printf() being a variadic function, no promotion (conversion) is performed. Otherwise, this is a well-defined behavior.

C11，章节 §7.21.6.1

C11, chapter §7.21.6.1

p 参数应该是一个指向 void 的指针.[……]

p The argument shall be a pointer to void. [.....]

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