Why are the outputs of printf and std::cout different?

Question

I tried the following code of C++. However, the outputs of printf and std::cout are different. Why?

struct Foo {     int a;     int b;     int c; };  int main() {     printf("%d\n", &Foo::c);  // The output is 8     std::cout << &Foo::c << "\n"; // The output is 1 } 

Answer 1

printf("%d\n", &Foo::c): this is undefined behavior, as &Foo::c is not an integer, but a pointer to member (but, actually, it is usual that the compiler stores pointer to data member as offset, and as 8 is the offset of Foo::c, 8 is printed).

std::cout << &Foo::c: this prints the value &Foo::c. As iostream doesn't have a pointer to member printer, it chooses the closest one: it converts it to bool, and prints it as integer. As &Foo::c converted to bool is true, 1 is printed.

Answer 2

The output is different because the behavior of your printf is undefined.

A pointer to member (like the one produced from &Foo::c) is not an integer. The printf function expects an integer, since you told it too with the %d specifier.

You can amend it by adding a cast to bool, like this:

printf("%d\n", (bool)&Foo::c) 

A pointer to member may be converted to a bool (which you do with the cast), and the bool then undergoes integral promotion to an int on account of being an integral variadic argument to a variadic function.

Speaking of the conversion to bool, it's exactly the conversion that is applied implicitly by attempting to call std::ostream's operator<<. Since there isn't an overload of the operator that supports pointers to members, overload resolution selects another that is callable after implicitly converting &Foo::c to a boolean.