How auto is deducing pointer type?

Question

In below code I could not understand how auto comes to know that thing on right hand side is pointer :

int x = 100;
int *ptr = & x;
auto test = ptr;
std::cout<<*test<<std::endl;

Because as per my understanding pointer contains address which is nothing but unsigned int so how auto deduces it to be pointer but not unsigned int?

Answer 1

you can also ask the question "eventually, everything in my program is bytes, so why does auto doesn't deduce everything to be uint8_t[]?"

Well, it's simple. the type of ptr is int* so the type of test is also int*. it doesn't matter how the generated assembly looks like. it may be that the cpu treats int* and unsigned int the same way, but that's irrelevant for C++, as C++ is a high level language.

besides that. the underlying statement that "a pointer is an unsigned int" is not true. pointer is a type that allows reading and writing to the memory address contained in that variable. an unsigned int is ... an unsigned int. nothing more, nothing less.

Answer 2

The standard says:

The type of a variable declared using auto is deduced from its initializer.

Thus, the type of test is deduced from the one of ptr (that is its initializer) and it's int *.

Note that auto follows almost the same rules of template type deduction.
You can refer to them for further details about the differences between auto, auto&, const auto &, auto&& and so on.