What's the difference between `auto pp` and `auto *ppp`? [duplicate]

This question already has an answer here:

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int foo = 11;
int *p = &foo;

auto pp = p;
auto *ppp = p;

cout << pp << endl;
cout << ppp << endl;

This program will produces the same output of pp and ppp,so why? auto deduce the variable should be int, so I think declaration of ppp is right. But pp and ppp have the same value... Output: 0x61fefc 0x61fefc