Kapil Kapil - 1 month ago 4
C++ Question

How auto is deducing pointer type?

In below code I could not understand how

comes to know that thing on right hand side is pointer :

int x = 100;
int *ptr = & x;
auto test = ptr;

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


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.