operator must take exactly one argument

Question

a h   include  logic h       class A   friend ostream amp  operator lt  lt  ostream amp   A amp             logic cpp   include  a h      ostream amp  logic  operator lt  lt  ostream amp  os  A amp  a                When i compile  it says      std  ostream amp  logic  operator lt  lt  std  ostream amp   A amp    must take exactly one argument    What is the problem

User · Accepted Answer

The problem is that you define it inside the class, which

a) means the second argument is implicit (this) and

b) it will not do what you want it do, namely extend std::ostream.

You have to define it as a free function:

class A { /* ... */ };
std::ostream& operator<<(std::ostream&, const A& a);

User · Answer

I ran into this problem with templated classes  Here s a more general solution I had to use   template class  lt T gt  class myClass       int myField          Helper function accessing my fields     void toString std  ostream amp   const          Friend means operator lt  lt  can use private variables        It needs to be declared as a template  but T is taken     template  lt class U gt      friend std  ostream amp  operator lt  lt  std  ostream amp   const myClass lt U gt   amp          Operator is a non-member and global  so it s not myClass lt U gt   operator lt  lt       Because of how C   implements templates the function must be    fully declared in the header for the linker to resolve it    template  lt class U gt  std  ostream amp  operator lt  lt  std  ostream amp  os  const myClass lt U gt   amp  obj      obj toString os     return os      Now    My toString   function can t be inline if it is going to be tucked away in cpp    You re stuck with some code in the header  I couldn t get rid of it    The operator will call the toString   method  it s not inlined   The body of operator lt  lt  can be declared in the friend clause or outside the class  Both options are ugly      Maybe I m misunderstanding or missing something  but just forward-declaring the operator template doesn t link in gcc   This works too   template class  lt T gt  class myClass       int myField          Helper function accessing my fields     void toString std  ostream amp   const          For some reason this requires using T  and not U as above     friend std  ostream amp  operator lt  lt  std  ostream amp   const myClass lt T gt   amp                 obj toString os           return os            I think you can also avoid the templating issues forcing declarations in headers  if you use a parent class that is not templated to implement operator lt  lt   and use a virtual toString   method

User · Answer

If you define operator lt  lt  as a member function it will have a different decomposed syntax than if you used a non-member operator lt  lt    A non-member operator lt  lt  is a binary operator  where a member operator lt  lt  is a unary operator      Declarations struct MyObj  std  ostream amp  operator lt  lt  std  ostream amp  os  const MyObj amp  myObj    struct MyObj          This is a member unary-operator  hence one argument     MyObj amp  operator lt  lt  std  ostream amp  os    os  lt  lt   this  return  this         int value   8         This is a non-member binary-operator  2 arguments std  ostream amp  operator lt  lt  std  ostream amp  os  const MyObj amp  myObj        return os  lt  lt  myObj value      So     how do you really call them   Operators are odd in some ways  I ll challenge you to write the operator lt  lt       syntax in your head to make things make sense   MyObj mo      Calling the unary operator mo  lt  lt  std  cout      which decomposes to    mo operator lt  lt  std  cout     Or you could attempt to call the non-member binary operator   MyObj mo      Calling the binary operator std  cout  lt  lt  mo      which decomposes to    operator lt  lt  std  cout  mo     You have no obligation to make these operators behave intuitively when you make them into member functions  you could define operator lt  lt  int  to left shift some member variable if you wanted to  understand that people may be a bit caught off guard  no matter how many comments you may write   Almost lastly  there may be times where both decompositions for an operator call are valid  you may get into trouble here and we ll defer that conversation   Lastly  note how odd it might be to write a unary member operator that is supposed to look like a binary operator  as you can make member operators virtual      also attempting to not devolve and run down this path       struct MyObj          Note that we now return the ostream     std  ostream amp  operator lt  lt  std  ostream amp  os    os  lt  lt   this  return os         int value   8       This syntax will irritate many coders now      MyObj mo   mo  lt  lt  std  cout  lt  lt   Words words words       this decomposes to    mo operator lt  lt  std  cout   lt  lt   Words words words           or even further     operator lt  lt  mo operator lt  lt  std  cout    Words words words      Note how the cout is the second argument in the chain here     odd right

User · Answer

A friend function is not a member function  so the problem is that you declare operator lt  lt  as a friend of A    friend ostream amp  operator lt  lt  ostream amp   A amp      then try to define it as a member function of the class logic   ostream amp  logic  operator lt  lt  ostream amp  os  A amp  a                      Are you confused about whether logic is a class or a namespace   The error is because you ve tried to define a member operator lt  lt  taking two arguments  which means it takes three arguments including the implicit this parameter  The operator can only take two arguments  so that when you write a  lt  lt  b the two arguments are a and b   You want to define ostream amp  operator lt  lt  ostream amp   const A amp   as a non-member function  definitely not as a member of logic since it has nothing to do with that class   std  ostream amp  operator lt  lt  std  ostream amp  os  const A amp  a      return os  lt  lt  a number

[c++] operator << must take exactly one argument

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