Hfrhyu

#include <iostream>

std::string get_data()

 return "Hello";


int main()

 const char* data = get_data().c_str();
 std::cout << data << "n";
 return 0;

"Hello" is printing on my machine; however, I am led to believe that this behavior is unspecified i.e. implementation-specific. Am I correct or will it always print "Hello", judging that the returned string is immutable and as such qualified as something that is constant?

The code exhibits undefined behavior.

get_data() returns a temporary which expires at the end of the full expression (*):

const char* data = get_data().c_str() ;
// ^~~~~~~~~~ ^
// this evaluates |
// to a prvalue |
// temporary expires here

data points to an internal of that object, so after the temporary ends you are left with a dangling pointer. Accessing it leads to Undefined Behavior. So the next line std::cout << data << "n"; makes the whole program exhibit Undefined Behavior.

*) There is an exception to this rule which doesn't apply here. If a prvalue is directly bound to a reference, the lifetime of the prvalue is extended to the lifetime of the reference.

For instance, this would have been fine:

int main()

 const std::string& ref = get_data();
 const char* data = ref.c_str();
 std::cout << data << "n";
 return 0;

Yes it is, but not the way you're doing it.

If you did this:

std::cout << get_data().c_str() << 'n';

you'd be just fine.

That's because a temporary is guaranteed to live for the lifetime of the full expression it was created in. It may live longer in certain, very specific circumstances.

If you bind a const reference to a temporary, it's lifetime will be extended to be the lifetime of the name it was bound to. So, code like this:

std::string const &x = get_data();
std::cout << x.c_str() << 'n';

would also work because the temporary returned by get_data would be bound to the reference named x, and so as long as x remained a valid name to use, the temporary would still exist.