Question

The following code works fine, but why is this correct code? Why is the “c_str()” pointer of the temporary returned by foo() valid? I thought, that this temporary is already destroyed when bar() is entered – but it doesn’t seem to be like this. So, now I assume that the temporary returned by foo() will be destroyed after the call to bar() – is this correct? And why?

std::string foo() {
  std::string out = something...;
  return out;
}

void bar( const char* ccp ) {
  // do something with the string..
}

bar( foo().c_str() );

Answer 1

$12.2/3- “Temporary objects are
destroyed as the last step in
evaluating the full-expression (1.9)
that (lexically) contains the point
where they were created. This is true
even if that evaluation ends in
throwing an exception.”

The lifetime of the temporary returned by foo() extends until the end of the full expression where it is created i.e. until the end of the function call ‘bar’.

EDIT 2:

$1.9/12- “A full-expression is an
expression that is not a subexpression
of another expression. If a language
construct is defined to produce an
implicit call of a function, a use of
the language construct is considered
to be an expression for the purposes
of this definition.”