Expression must have class type

Question

I have t coded in c   for some time and I got stuck when I tried to compile this simple snippet   class A     public      void f          int main             A a      a f       works fine              A  a   new A        a f       this doesn t

User · Answer

a is a pointer  You need to use- gt   not

User · Answer

Allow an analysis    include  lt iostream gt       not  include  iostream  using namespace std      in this case okay  but never do that in header files  class A    public    void f     cout lt  lt  f   n         int main             A a    this works  A  a   new A      this doesn t  a f        f has not been declared         below       system  pause      lt -- Don t do this  It is non-portable code  I guess your                            teacher told you this                            Better  In your IDE there is prolly an option somewhere                                   to not close the terminal console-window                            If you compile on a CLI  it is not needed at all      As a general advice    0  Prefer automatic variables   int a    MyClass myInstance    std  vector lt int gt  myIntVector   1  If you need data sharing on big objects down     the call hierarchy  prefer references     void foo  std  vector lt int gt  const  amp input          void bar              std  vector lt int gt  something                    foo  something         2  If you need data sharing up the call hierarchy  prefer smart-pointers    that automatically manage deletion and reference counting   3  If you need an array  use std  vector lt  gt  instead in most cases     std  vector lt  gt  is ought to be the one default container   4  I ve yet to find a good reason for blank pointers      - gt  Hard to get right exception safe         class Foo              Foo      a new int 512    b new int 512                  Foo                    delete    b                 delete    a                                 - gt  if the second new   fails  Foo leaks memory  because the           destructor is never called  Avoid this easily by using            one of the standard containers  like std  vector  or           smart-pointers    As a rule of thumb  If you need to manage memory on your own  there is generally a superiour manager or alternative available already  one that follows the RAII principle

User · Answer

It s a pointer  so instead try   a- gt f      Basically the operator    used to access an object s fields and methods  is used on objects and references  so   A a  a f    A amp  ref   a  ref f      If you have a pointer type  you have to dereference it first to obtain a reference   A  ptr   new A      ptr  f    ptr- gt f      The a- gt b notation is usually just a shorthand for   a  b   A note on smart pointers  The operator- gt  can be overloaded  which is notably used by smart pointers  When you re using smart pointers  then you also use - gt  to refer to the pointed object   auto ptr   make unique lt A gt     ptr- gt f

User · Answer

Summary  Instead of a f    it should be a- gt f     In main you have defined a as a pointer to object of A  so you can access functions using the - gt  operator   An alternate  but less readable way is   a  f    a f   could have been used to access f    if a was declared as  A a

[c++] Expression must have class type

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