Ampersands

What are ampersands for when used with member declarations? I first noticed it here: [15] Input/output via <iostream> and <cstdio>, C++ FAQ

Code:

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#include <iostream>
 
 class Fred {
 public:
  friend std::ostream& operator<< (std::ostream& o, Fred const& fred);
  ...
 private:
  int i_;    // Just for illustration
 };
 
 std::ostream& operator<< (std::ostream& o, Fred const& fred)
 {
  return o << fred.i_;
 }
 
 int main()
 {
  Fred f;
  std::cout << "My Fred object: " << f << "\n";
  ...
 }

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std::ostream&, what's it for? And what's the dif in effect between pre and post fix.

It specifies that the return type is a reference to a std::ostream.

Quote:

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Originally Posted by User Name:

And what's the dif in effect between pre and post fix.

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In what context?

Code:

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template<class T>
T &ByteArray::operator[](unsigned int i)
{
        return (T(bytes))[i];
}

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Well (T(bytes))[i++], (T(bytes))[++i], (T(bytes)[i]++, and ++(T(bytes))[i] are certainly all different. The difference lies both in what they increase and in what they return. Postfix increment will return the variable's value as it was, before the increment. In prefix, the variable is incremented and then return the value. Increments bind to the thing they are closest to, so some of the statements like i++ increase i, and the others increase the bytes element. Knowing the difference could save some little gray cells.

I'm aware of how pre/post increment works. I was asking about what the ampersand("&") does when pre/suf-fixed to a method, and what the difference, in effect, between prefix and suffix.

There is no prefix and postfix for &, which are called references. They appear next to a type and specifies a reference.
A reference is basically an alias for an existing variable. Sort of like pointers. Google references for more information.

There is no difference, just as there is no difference where you put the * with pointers.

Code:

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char* foo();
char *bar();

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