Call of overloaded function is ambiguous

Question

What does this error message mean   error  call of overloaded    setval int     is ambiguous huge cpp 18  note  candidates are  void huge  setval unsigned int  huge cpp 28  note                  void huge  setval const char     My code looks like this    include  lt iostream gt   define BYTES 8 using namespace std    class huge   private      unsigned char data BYTES   public      void setval unsigned int       void setval const char           void huge  setval unsigned int t        for int i   0  i lt  BYTES   i              data i    t          t   t  gt  gt  1           void huge  setval const char  s        for int i   0  i lt  BYTES   i            data i    s i      int main         huge p      p setval 0       return 0

User · Answer

Cast the value so the compiler knows which function to call:

p.setval(static_cast<const char *>( 0 ));

Note, that you have a segmentation fault in your code after you get it to compile (depending on which function you really wanted to call).

User · Answer

The literal 0 has two meanings in C    On the one hand  it is an integer with the value 0  On the other hand  it is a null-pointer constant   As your setval function can accept either an int or a char   the compiler can not decide which overload you meant   The easiest solution is to just cast the 0 to the right type  Another option is to ensure the int overload is preferred  for example by making the other one a template   class huge    private    unsigned char data BYTES    public    void setval unsigned int     template  lt class T gt  void setval const T        not implemented   template  lt  gt  void setval const char

User · Answer

Use   p setval static cast lt const char   gt  0      or  p setval static cast lt unsigned int gt  0      As indicated by the error  the type of 0 is int   This can just as easily be cast to an unsigned int or a const char     By making the cast manually  you are telling the compiler which overload you want

User · Answer

That is ambiguous because a pointer is just an address  so an int can also be treated as a pointer     0  an int  can be converted to unsigned int or char   equally easily  The short answer is to call p setval   with something that s unambiguously one of the types it s implemented for  unsigned int or char     p setval 0U   p setval  unsigned int 0   and p setval  char   0  will all compile  It s generally a good idea to stay out of this situation in the first place  though  by not defining overloaded functions with such similar types

User · Answer

The solution is very simple if we consider the type of the constant value  which should be  unsigned int  instead of  int    Instead of   setval 0    Use   setval 0u    The suffix  u  tell the compiler this is a unsigned integer  Then  no conversion would be needed  and the call will be unambiguous

User · Answer

replace p setval 0   with the following   const unsigned int param   0  p setval param     That way it knows for sure which type the constant 0 is

[c++] Call of overloaded function is ambiguous

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