why are the addresses of these variables printing as ab@ and b@?
c++, memory-address
Solution
When you write
cout << &test.a
because `test.a` is a `char`, this will invoke the `operator<< (ostream&, const char*)` overload, thinking that your `char*` is a pointer to a C-style string rather than a pointer to just one character. Consequently, the operation will start reading bytes starting at the memory address of `&test.a` until it finds a null terminator (a zero byte). This happens to print out `ab@`, since `a` has value `'a'`, `b` has value `'b'`, and the number 478, on your system, happens to correspond to an `@` character followed eventually by a null byte.
If you want to see the numeric addresses of `test.a` and `test.b`, cast the pointers to `void*`s, which will select the `operator<< (ostream&, const void*)` overload. For example:
cout << static_cast<void*>(&test.a)
Hope this helps!
Problem
I run the follow code: ``` #include <iostream> using namespace std; typedef struct Test { char a; char b; int i; double d; }Test; int main() { Test test; test.a = 'a'; test.b = 'b'; test.i = 478; test.d = 4.7; cout << &test.a << '\n' << &test.b << '\n' << &test.i << '\n' << &test.d << '\n'; return 0; } ``` The output is: ``` ab@ b@ 0x28fe94 0x28fe98 ``` At first, i thought it is a result of the precedence between `&` and `.`. But the `0x28fe94` and `0x28fe94` indicate it's not the problem of precedence. I can figure out what does the `ab@` and `b@` mean?