How can I Replace raw pointers with Smart pointers?

Question

specifically when work with some functions as belong to external library.
It will be very clear if it presented as comprehensive example.

Answer 1

Just replace "T* p = new T(...);"
with "std::unique_ptr<T> p( new T(...) );" // or shared_ptr
or "p.reset( new T(...) );" after initialization
and functions that expect a T* simply use "p.get()".
The rest is pretty much the same as a raw pointer.

For example:

void f( int* p ){ *p = 4; }

int main()
{
int* p = new int( 5 );
f( p );
std::cout << *p;
delete p;
}

Becomes:

void f( int* p ){ *p = 4; }

int main()
{
std::unique_ptr<int> p( new int( 5 ) ); // Don't use =
f( p.get() );
std::cout << *p;
}

Answer 2

You'd put them into a separate variable and then get that pointer's address like.
"
std::unique_ptr<int> p;
int* p_tmp = p.get();
foo( &p_tmp, ptr_d.get() );
"

If foo assigns a new address you'd have to update the unique_ptr too with
"p.reset( p_tmp );"
Maybe even with
"p.reset( std::exchange( p_tmp, nullptr ) );"
to cleanup p_tmp as well by setting it to nullptr if the object is move constructible.

Not sure if there are better ways. You probably want to avoid ** as much as possible. :)

Answer 3

Thanks Dennis.
And one more thing, what about pointer to pointer arguments?

void foo(int** ptr, double* ptr_d) ;

according to your answer preceding statement convert to:

unique_ptr <int> ptr_d { make_unique <int>(9)} ;

void (???, ptr_d. get()) ;