c++ - How is ambiguity determined in the overload resolution algorithm? -

I am trying to understand the overloading resolution method.

Why is it unclear:

Void func (int, double, double, double) {} void main () {func (1, 2, 3, 4);>

  Void func (double, int, int, double) {}}   
 But is not it? Zero (int, double, double, double) {} zero main () {throw (1, 2, 3, 4) 
 
  zero (int, int, int, double) vic ; }  
 
 In the first case, there are 2 precise parameter matches and 2 conversions against 1 exact match and 3 conversions, and in the second case there are 3 exact matches and 1 conversion with 1 exact match And 3 conversions 
 
 So why is not one unclear and one? What is the logic here? Overload resolution rules define partial orders only on all sets of matches - if an overload  F1 is not a good match with   F2 , this does not mean that  F2  is this  F1 . The exact partial order can be considered about comparing two points in the  k  dimension, where the number of logic is  k . Define this partial order on the number in  k  -dim space -  (x_1, x_2, ..., x_k). (Y_1, y_2, ..., y_k) if x_i for all i and x_j & lt; = Y_i & lt; For at least one J  y_j this is actually a partial order on non-temp function candidates defined by standard. 
 
 View your instances: 
 
  Zero function (double, int, int, double) {Vvv vvv vvv Better better equal zero func (int, double, double, Double} {VVV VVV Better Than  
 
 Therefore the surcharge is strictly better than the other. 
 
 In your second example: 
 
  zero func (int, int, int, double) {} vvv vvv vvv vvv equals better better zero function (int, double, Double, double)) Vvv equal  
 
 Now, the first surcharge is better than the other one at all, but there is a logic and not worse than the other. Thus, there is no ambiguity. - The partial order actually declares the first one better. 
 
 (The above description function does not consider the template. You can find more information.)