商店COUT

Question

I的值具有下面的代碼

#include <iostream> 
#include <math.h> 
#include <string> 
#include <sstream> 
using namespace std; 
int main() { 
int value 
cout << "Input your number: " << endl; 
cin >> value; 
string s = to_string(value); 
const int count = s.length(); 
int position = count; 
for (int i = 1; i < count + 1; i++) 
{ 
    int pwr = pow(10, position - 1); 
    cout << ((value/pwr) + position) % 10; 
    position--; 
    value = value % pwr; 
} 

相反COUT的，我怎樣才能使用for循環的((value/pwr) + position) % 10值存儲到一個變量。非常感謝你的幫助。

[編輯] 我增加了一個陣列，而不是

int val[7]; 
int position = count; 
for (int i = 1; i < count + 1; i++) 
{ 
    int pwr = pow(10, position - 1); 
    val[i-1] = ((value/pwr) + position) % 10; 
    position--; 
    value = value % pwr; 
} 
cout << "Encoded value is: "; 
for (int i = 0; i < 8; i++) 
{ 
    cout << val[i]; 
} 

它能夠輸出I想要的值，但有一個運行時故障＃2 - 疊圍繞變量「VAL」被損壞。這是爲什麼？

Answer 1

您可以使用像矢量一樣的STL容器。

#include <iostream> 
#include <math.h> 
#include <string> 
#include <sstream> 
#include <vector> 

using namespace std; 
int main() { 
    int value; 
    cout << "Input your number: " << endl; 
    cin >> value; 
    string s = to_string(value); 
    const int count = s.length(); 
    int position = count; 

    vector<int> outputs; 

    for (int i = 1; i < count + 1; i++) 
    { 
    int pwr = pow(10, position - 1); 
    outputs.push_back(((value/pwr) + position) % 10); 
    cout << outputs.back(); 

    position--; 
    value = value % pwr; 
    } 

    cout << endl; 

    // iterate through the vector later 
    for (auto i : outputs) 
    { 
    cout << i; 
    } 

    cin.get(); 
} 

Answer 2

商店載體，然後打印

它不是那麼難。至於你不知道會有怎樣的規模或如何結果數多將在那裏它是使用矢量有用。

此處參考http://en.cppreference.com/w/cpp/container/vector

#include <iostream> 
#include <math.h> 
#include <string> 
#include <sstream> 
#include<vector> 
#include<iterator> 
#include<algorithm> 

using namespace std; 

int main() { 
    int value; 
    vector<int> results; 
    cout << "Input your number: " << endl; 
    cin >> value; 
    string s = to_string(value); 
    const int count = s.length(); 
    int position = count; 
    for (int i = 1; i < count + 1; i++) 
    { 
     int pwr = pow(10, position - 1); 
     auto val = ((value/pwr) + position) % 10; 
     results.push_back(val); 
     position--; 
     value = value % pwr; 
    } 
    copy(results.begin(),results.end(),ostream_iterator<int>(cout," ")); 

} 

輸出

Input your number: 
12 
3 3 Program ended with exit code: 0