Call of overloaded function is ambiguous

问题:

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 <iostream>
#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< BYTES ; i++) {
        data[i] = t;
        t = t >> 1;
    }
}

void huge::setval(const char *s) {
    for(int i = 0; i< BYTES ; i++)
        data[i] = s[i];
}

int main() {
    huge p;
    p.setval(0);
    return 0;
}

回答1:

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 <class T> void setval(const T *); // not implemented
  template <> void setval(const char*);
};

回答2:

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.

回答3:

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.

回答4:

Use

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

or

p.setval(static_cast<unsigned int>(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.

回答5:

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).

回答6:

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.