Why is this conversion happening?

#include<iostream>

using namespace std;

class test
{
  int a, b;
public开发者_Go百科:

  test() {
    a=4; b=5;
  }

  test(int i,int j=0) {
    a=i; b=j;
  }

  test operator +(test c) {
     test temp;
     temp.a=a+c.a;
     temp.b=b+c.b;
     return temp;
  }

  void print() {
    cout << a << b;
  }
};

int main() {
  test t1, t2(2,3), t3;
  t3 = t1+2;
  t3.print();
  return 0;
}

How can the compiler accept a statement like t3=t1+2; where 2 is not an object?

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The compiler sees you are invoking operator+(test) and attempts to implicitly convert the 2 to a test successfully using your test(int i,int j=0) constructor.

If you want to make the conversion explicit, you must change the constructor to explicit test(int i, int j=0). In this case, your code would generate a compiler error because 2 cannot be implicitly converted to test. You would need to change the expression to t1 + test(2).

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Because test(int i,int j=0) is a constructor that takes one or two arguments, so a test object is created from 2. In the next stage test operator +(test c) is called.

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There is a binary operator+ available which takes two operands of type test. Moreover, test is implicitly constructible from int via the constructor test(int, int = 0). Putting the two together, t1 + 2 becomes t1 + test(2, 0).

To disallow this silent conversion (which sometimes can cause very surprising conversion chains), declare your constructors that accept one single argument as explicit: explicit test(int, int = 0).

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Because test(int i, int j = 0) is not marked explicit.

Therefore t1 + 2 is interpreted as t1.operator+(2) which is itself interpreted as t1.operator+(test(2)) (implicit conversion).

If you mark the constructor as explicit, an error will occur (during the compilation), saying that 2 cannot be converted into a test or that the operator+ does not match.

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In short, because C++ includes operator overloading, the ability to define custom implementations of operators for user-defined types. The operator+() function shown above is how you define the + operator for the type test. When the compiler sees an expression in which + is applied to a test object, it looks for an operator+ defined in test (or a two-argument form defined as a global function with a first argument of type test or test&.) It then invokes the function, possibly after converting the other argument.

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Because your constructor test(int i,int j=0) introduces a user defined conversion from int to test.

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The compiler creates an object of type Test using its constructor.

 t3 = t1 + test(2);