如何應用循環前提條件我有點困惑

Question

我是一個初學者，所以請裸露在我這是我的代碼我必須得到無限的輸入，但是當兩個最近的一個相等的15之間的差異輸出這兩個數字。

#include <iostream> 
#include <cstdlib> 
using namespace std; 
bool DiffCheck(int x, int y); 
int main(int argc, const char * argv[]) { 
int latest, prior; 
cout << "Enter some integers:\n"; 
cin >> prior; 
cin >> latest; 
while (DiffCheck(latest, prior)) { 


prior = latest; 
cin >> latest; 

}

cout << "The difference between " << prior << " and " << latest << " is greater than 15" << endl; 
return 0; 
} 
bool DiffCheck(int x, int y) 
{ 
return(abs(x - y) >= 15) ? false : true; 
} 

Answer 1

#include <iostream> 

using namespace std; 
int main() 
{ 
    int first = 0; 
    int second = 0; 
    while(1) 
    { 
    cin>>second; 

    if(second - first >= 15) 
     { 
      cout<<"First: "<<first<<"Second: "<<second<<endl; 
      break; // stop the code once the difference is 15 or greater 
     } 

    first = second; 
    } 

    return 0; 
} 

注：用戶正在尋找一個代碼，如果數量之間的差爲15或更高，將停止。所以，改變了代碼。