Is using the address of an uninitialized variable UB? [duplicate]

问题

Is this small code UB?

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

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

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

回答1:

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

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. :) )

That said, strictly speaking, you should write

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

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, chapter §7.21.6.1

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

回答2:

Is this small code UB?

Yes, it's UB because the conversion specifier p requires a void-pointer.

On the other hand the code below does not invoke UB

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

as the address of bar is well defined independently whether bar itself got initialised.

回答3:

This behavior is well defined.

The address of the variable is known. The fact that it hasn't been explicitly initialized doesn't matter.

来源：https://stackoverflow.com/questions/39918506/is-using-the-address-of-an-uninitialized-variable-ub

标签

printf

undefined-behavior

variadic-functions