Overloading operators in typedef structs c

Question

I want to make a typedef struct called pos  from position  that stores coordinates x and y  I am trying to overload some operators for this struct  but it does not compile   typedef struct       int x      int y       inline pos operator  pos a            x a x          y a y          return a             inline pos operator  pos a            return  a x x a y y              inline bool operator   pos a           if  a x  x  amp  amp  a y   y            return true         else           return false          pos    I also wanted to know the difference between this   inline bool operator   pos a        if a x  x  amp  amp  a y   y         return true        else        return false      And this   bool operator   pos a  const         if a x  x  amp  amp  a y   y           return true        else          return false

User · Answer

try this   struct Pos      int x      int y       inline Pos amp  operator  const Pos amp  other           x other x          y other y          return  this             inline Pos operator  const Pos amp  other  const           Pos res  x other x y other y           return res             const inline bool operator   const Pos amp  other  const           return  x  other x and y    other y

User · Answer

Instead of typedef struct         pos  you should be doing struct pos            The issue here is that you are using the pos type name before it is defined   By moving the name to the top of the struct definition  you are able to use that name within the struct definition itself   Further  the typedef struct         name  pattern is a C-ism  and doesn t have much place in C     To answer your question about inline  there is no difference in this case   When a method is defined within the struct class definition  it is implicitly declared inline   When you explicitly specify inline  the compiler effectively ignores it because the method is already declared inline    inline methods will not trigger a linker error if the same method is defined in multiple object files  the linker will simply ignore all but one of them  assuming that they are all the same implementation   This is the only guaranteed change in behavior with inline methods   Nowadays  they do not affect the compiler s decision regarding whether or not to inline functions  they simply facilitate making the function implementation available in all translation units  which gives the compiler the option to inline the function  if it decides it would be beneficial to do so

User · Answer

The breakdown of your declaration and its members is somewhat littered   Remove the typedef  The typedef is neither required  not desired for class struct declarations in C    Your members have no knowledge of the declaration of pos as-written  which is core to your current compilation failure   Change this   typedef struct        pos    To this   struct pos              Remove extraneous inlines  You re both declaring and defining your member operators within the class definition itself  The inline keyword is not needed so long as your implementations remain in their current location  the class definition     Return references to  this where appropriate   This is related to an abundance of copy-constructions within your implementation that should not be done without a strong reason for doing so  It is related to the expression ideology of the following   a   b   c    This assigns c to b  and the resulting value b is then assigned to a  This is not equivalent to the following code  contrary to what you may think   a   c  b   c    Therefore  your assignment operator should be implemented as such   pos amp  operator   const pos amp  a        x   a x      y   a y      return  this      Even here  this is not needed  The default copy-assignment operator will do the above for you free of charge  and code  woot    Note  there are times where the above should be avoided in favor of the copy swap idiom  Though not needed for this specific case  it may look like this   pos amp  operator  pos a     by-value param invokes class copy-ctor       this- gt swap a       return  this      Then a swap method is implemented   void pos  swap pos amp  obj           TODO  swap object guts with obj     You do this to utilize the class copy-ctor to make a copy  then utilize exception-safe swapping to perform the exchange  The result is the incoming copy departs  and destroys  your object s old guts  while your object assumes ownership of there s  Read more the copy swap idiom here  along with the pros and cons therein     Pass objects by const reference when appropriate  All of your input parameters to all of your members are currently making copies of whatever is being passed at invoke  While it may be trivial for code like this  it can be very expensive for larger object types  An exampleis given here   Change this   bool operator   pos a  const      if a x  x  amp  amp  a y   y return true      else return false      To this   also simplified   bool operator   const pos amp  a  const       return  x    a x  amp  amp  y    a y       No copies of anything are made  resulting in more efficient code     Finally  in answering your question  what is the difference between a member function or operator declared as const and one that is not    A const member declares that invoking that member will not modifying the underlying object  mutable declarations not withstanding   Only const member functions can be invoked against const objects  or const references and pointers  For example  your operator     does not modify your local object and thus should be declared as const  Your operator     clearly modifies the local object  and therefore the operator should not be const     Summary  struct pos       int x      int y          default   parameterized constructor     pos int x 0  int y 0             x x   y y                      assignment operator modifies object  therefore non-const     pos amp  operator  const pos amp  a                x a x          y a y          return  this                addop  doesn t modify object  therefore const      pos operator  const pos amp  a  const               return pos a x x  a y y                 equality comparison  doesn t modify object  therefore const      bool operator   const pos amp  a  const               return  x    a x  amp  amp  y    a y                EDIT OP wanted to see how an assignment operator chain works  The following demonstrates how this   a   b   c    Is equivalent to this   b   c  a   b    And that this does not always equate to this   a   c  b   c    Sample code    include  lt iostream gt   include  lt string gt  using namespace std   struct obj       std  string name      int value       obj const std  string amp  name  int value            name name   value value                   obj amp  operator   const obj amp  o                cout  lt  lt  name  lt  lt         lt  lt  o name  lt  lt  endl          value    o value 1      note  our value is one more than the rhs          return  this                int main int argc  char  argv           obj a  a   1   b  b   2   c  c   3        a   b   c      cout  lt  lt   a value      lt  lt  a value  lt  lt  endl      cout  lt  lt   b value      lt  lt  b value  lt  lt  endl      cout  lt  lt   c value      lt  lt  c value  lt  lt  endl       a   c      b   c      cout  lt  lt   a value      lt  lt  a value  lt  lt  endl      cout  lt  lt   b value      lt  lt  b value  lt  lt  endl      cout  lt  lt   c value      lt  lt  c value  lt  lt  endl       return 0      Output  b   c a   b a value   5 b value   4 c value   3 a   c b   c a value   4 b value   4 c value   3

User · Answer

bool operator   pos a  const  - this method doesn t change object s elements  bool operator   pos a    - it may change object s elements

[c++] Overloading operators in typedef structs (c++)

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