Why does std::cout convert volatile pointers to bool?

Question

If you try to cout a pointer to a volatile type, even a volatile char pointer where you would normally expect cout to print the string, you will instead simply get \'1\' (as

Answer 1

I think the reason is that volatile pointers cannot be converted implicitly to void *. This is in Appendix C of the Standard, and the rationale is type safety.

Change: Only pointers to non-const and non-volatile objects may be implicitly converted to void* Rationale: This improves type safety.

So instead of the conversion to void * (which would print in hex), you get the "default" conversion to bool.