A ++ B, anb

A ++ B, anb
One of my questions:First, let's take a look at the function implementation of the following class. ++ indicates that each bit of the class's data is + 1. If it is not carried, an integer (the processed data) is returned ), + indicates the addition of two classes. each bit of the class is added, and the result is an integer. Code implementation: testmain. cpp:

#include <iostream>#include "strange_plus.h"using namespace std;int main() {    integer a, b;    int t1, t2, ans;    t1 = a + b;    t2 = ++a;    cout << t1 << endl;    cout << t2 << endl;    ans = a+(++b);    cout << ans << endl;    return 0;}

Strange_plus.h:

#include <string.h>class integer {  public:    integer() {        data = 1;    }    ~integer() {}        friend int operator + (const integer i1, const integer i2) {        int temp1 = i1.data;        int temp2 = i2.data;        int temp3 = 0;        int ans = 0;        while (temp1 || temp2) {            temp3 = temp3 * 10 + ((temp1 % 10 + temp2 % 10) % 10);            temp1 /= 10;            temp2 /= 10;        }        while (temp3) {            ans = ans * 10 + temp3 % 10;            temp3 /= 10;        }        return ans;    }        int operator ++ () {        int temp[10];        memset(temp, 0, sizeof(temp));        int temp_int = data;        int i = 0;        while (temp_int) {            temp[i++] = temp_int % 10;            temp_int /= 10;        }        for (int j = 0; j < i; j++) {            temp[j] = (temp[j] + 1) % 10;        }        int ans = 0;        for (int j = i - 1; j >= 0; j--) {            ans = ans * 10 + temp[j];        }        data = ans;        return ans;    }      private:    int data;};

Then an error is reported. Check the error message:
I don't know much about it, but I found it quite strange. If I add this: integer (int temp) {data = temp;} In strange_plus.h, I can run it:
The analysis is as follows: first, for a + (++ B), calculate ++ B and return an integer. For this integer, it cannot be added to the class, after adding the code integer (int temp) {data = temp;}, you can convert the integer into a class and add it together;
In fact, this phenomenon is the static binding rule of the function, as mentioned earlier:
Here is the third step. In this case, the constructor integer (int temp) {data = temp;} is equivalent to forced conversion. Similarly, if ++ is changed to the Postfix, a + (++ B) is changed to a ++ B. Note the priority: the front-end ++> front-end ++ has the same priority as the front-end ++, but is lower than the back-end ++.