meaning of *& syntax

[Hansen]

Hi there,

I just came across the following syntax in a class declaration:

my_base*& base_;

What does that mean? Is it a pointer to a reference? In my head * and & are
opposites meaning that my_base*& == my_base ...
But I have a feeling that I'm wrong ... )

Best Regards
Hansen

 

[Rolf Magnus]

Hi there,

I just came across the following syntax in a class declaration:

my_base*& base_;

What does that mean? Is it a pointer to a reference?

No. There are no such things as pointers to references. It's just the other
way round. It's a reference to a pointer.

In my head * and & are opposites meaning that my_base*& == my_base ...

my_base*& is a type. What you mean would be *&my_base.

 

[Clark S. Cox III]

Hi there,

I just came across the following syntax in a class declaration:

my_base*& base_;

What does that mean? Is it a pointer to a reference?

Close, it's a reference to a pointer.

In my head * and & are
opposites meaning that my_base*& == my_base ...
But I have a feeling that I'm wrong ... )

Just like (int&) is a reference to an int, (int*&) is a reference to an
int pointer.

 

[Grizlyk]

I just came across the following syntax in a class declaration:

my_base*& base_;

What does that mean?

"cdecl/cppdecl" can help you sometimes to resolve c-language expressions.

Also find a table with "C++ precedens of operators": all operators have
divided on groups by "priority level" and "associativity" ("left to right"
or "right to left").

 

[Clark S. Cox III]

"cdecl/cppdecl" can help you sometimes to resolve c-language expressions.

Also find a table with "C++ precedens of operators": all operators have
divided on groups by "priority level" and "associativity" ("left to right"
or "right to left").

This is not entirely true, some of the relationships between the
operators cannot be expressed in a simple precedence table.

 

[Grizlyk]

This is not entirely true, some of the relationships between the
operators cannot be expressed in a simple precedence table.

For example?

 

[=?ISO-8859-15?Q?Juli=E1n?= Albo]

For example?

Yes.

 

[Clark S. Cox III]

For example?

Consider the relationship between the '=' operator and the '?:' operator:

a?b:c=3; //Equivalent to a?bc=3)
a?b=3:c; //Equivalent to a?(b=3):c
a=3?b:c; //Equivalent to a=(3?b:c)

 

[Craig Scott]

I just came across the following syntax in a class declaration:

Close, it's a reference to a pointer.


Just like (int&) is a reference to an int, (int*&) is a reference to an
int pointer.

Since the OP hasn't seen this before, I mention the following as
additional info. It can be useful to pass a "reference to a pointer"
to functions that need to change what the pointer points to from the
*caller's* point of view. For example (this is a very contrived
example):

void func(int*& foo, int* t)
{
// A real function would obviously do something
// more meaningful here!
foo = t;
}

int main(int argc, char* argv[])
{
int a = 0;
int b = 1;
int* myFoo = &a;

// This prints the value of a, which is 0
std::cout << *myFoo << std::endl;

// This will change myFoo to point to b
func(myFoo, &b);

// This now prints the value of b, which is 1
std::cout << *myFoo << std::endl;
}


Member variables can also be used in a similar fashion, but in my
experience, that is less common (but not any more or less valid).