The output of function overloading is not as expected

Question

I'm learning about function overloading in C++ and I was having a doubt on function matching. Running below code I was getting error C2668: 'function' : ambiguous call to overloaded function.

The reson why I got error was clearly answered in this link. The number 1.2 and 2.2 are type double.Although there is one more doubt which I have and was not answered there. When I try to call function(1.2,2) or function(1,2.2) it is printing as "int function". Why it is not giving error same as above.

CODE:

  void function(int y,int w)
    {
      printf("int function");

    }


 void function(float y,float w)
  {
    printf("float function");
  }


int main()
 {
   function(1.2,2.2);
   return 0;
 }

@πάνταῥεῖ: He _has_ provided an MCVE, it's just at the link. Suggest reopening. — einpoklum, Nov 25 '18 at 09:16
@YePhIcK: Did you follow his link? It has all the necessary context IMHO. — einpoklum, Nov 25 '18 at 09:16
@user4156958: You should copy code from the linked-to question to satisfy our friends here.. — einpoklum, Nov 25 '18 at 09:20
@user4156958 You should edit your question as mentiond please. — πάντα ῥεῖ, Nov 25 '18 at 09:20
Anyway, my answer (which I would post as an answer if I could) is: the calls you listed are with a double and an int or an int and a double, which requires just _one_ casting to get to the function(int, int) signature, but two casts to get to the function(float, float) signature. That's probably why the former is preferred over the latter. — einpoklum, Nov 25 '18 at 09:21
@einpoklum A question **has to be** self-contained. This implies code has to be **in the question**, not linked. I recommend reading [ask] and have a look at the close reasons. — too honest for this site, Nov 25 '18 at 09:27
"What is overload resolution?" in [AlokSave's answer](https://stackoverflow.com/a/16602198/3975177) on the question the OP linked answers OPs question perfectly. — Swordfish, Nov 25 '18 at 10:19
1.2 is a literal of type double, not float. So a conversion is required and it can go either to int or double. They are both lossy conversions that lose precision, the compiler isn't convinced one is better than the other. Use 1.2f — Hans Passant, Nov 25 '18 at 10:43
I know that.That is the reason I'm getting error.But when I'm giving argument like **function(1.2, 2 )** it is giving output without any error. @HansPassant — Yogesh Tripathi, Nov 25 '18 at 10:47
When you use an int literal then the compiler can find a function candidate that requires only one conversion instead of two. A better match and there's only one of them. So no ambiguity anymore about which one it is better and it is happy to coerce double to int. — Hans Passant, Nov 25 '18 at 10:51

score 1 · Answer 1 · answered Nov 25 '18 at 18:22

When I try to call function(1.2,2) or function(1,2.2) it is printing as "int function". Why it is not giving error same as above.

Because according to the rules of overload resolution void function(int y,int w) is a better match than void function(float y,float w) and therefore there is no ambiguity.

Calls function(1.2,2) and function(1,2.2) both have one argument that is an int and is the exact match with one of the arguments of void function(int y,int w), and so only one type conversion is required, double to int. void function(float y,float w) on the other hand requires two conversions int to float and double to float, and this is why int overload is used.

The output of function overloading is not as expected

1 Answers1